Chapter – 2 New Kings And Kingdoms | Class 7th | NCERT History Solutions | Edugrown

NCERT Solutions for Class 7 Social Science includes all the questions provided in NCERT Class 7 Social Science Text Book of Geography The Earth: Our Habitat, History Our Pasts, Civics Social and Political Life. Here CBSE Class 7 SST all questions are solved with the detailed explanation to score good marks in the exams. you can check Extra Questions for Class 7 Social Science

Chapter -2 New Kings And Kingdoms

1. Match the following:
Gwjara-Pratiharas    –      Western Deccan
Rashtrakutas             –       Bengal
Palos                           –     Gujarat and Rajasthan
Cholas                         –        Tamil Nadu
Answer: 
Gurj ara-Pratiharas        –   Gujarat and Rajasthan
Rashtrakutas                   –   Western Deccan
Palas                                 –    Bengal
Cholas                               –   Tamil Nadu

2. Who were the parties involved in the ‘tripartite struggle’?
Answer: The parties involved in the tripartite struggle were Gurjar-Pratihara, Rashtrakuta, and Pala dynasties.

3. What were the qualifications necessary to become a member of a committee of the sabha in the Chola Empire?
Answer:
The working of a sabha according to the Uttaramerur inscription is given below:

  1.  All those who wish to become members of the sabha should be owners of the land from which land, revenue is collected.
  2. They should have their own homes.
  3. They should be between 35 and 70 years of age.
  4. They should have knowledge of the Vedas.
  5. They should be well-versed in administrative matters and honest in all dealings.
  6. If anyone has been a member of any committee in the last three years, he cannot become a member of another committee.
  7. Anyone who has not submitted his accounts, as well as those of his relatives, cannot contest the elections.

4. What were the two major cities under the control of the Chahamanas?
Answer:  
Delhi and Ajmer were the two cities under control of the Chahamanas.

5. How did the Rashtrakutas become powerful?
Answer:
Rashtrakutas became powerful in the following manner:

  1. Rashtrakutas were subordinates to the Chalukyas of Karnataka.
  2. In the mid-eighth century Dantidurga, a Rashtrakuta chief, overthrew his Chalukyan overlord.
  3. He performed a ritual called Hiranya-garbha with the help of Brahmanas.
  4. This ritual was, then, considered to lead the rebirth of the sacrificer as Kshatriya, even if he was not Kshatriya by birth.

6. What did the new dynasties do to gain acceptance?
Answer:  The new dynasties gained power and wealth. Thereafter they declared themselves to be maha-samantas or mahamandaleshwara. Many of such kings adopted high sounding titles like maharaja-adhiraja or tribhuvana-chakravartin. They also deputed learned brahmanas to depict them as valiant, victorious warriors. Their activities were recorded in Prashastis. They tried to demonstrate their power and resources by building large temples.

7. What kind of irrigation works were developed in the Tamil Nadu?
Answer:
Large scale cultivation was developed in Tamil Nadu and in many cases artificial irrigation became necessary. A variety of methods were used:

  • Wells were dug
  • Huge tanks were constructed to collect rainwater
  • Canals were constructed

All this work required planning, organizing labour and resources. Decisions were taken collectively on the usage of water. New rulers and people took an active interest in these activities.

8. What were the activities associated with Chola temples?
Answer:
Splendid temples and bronze sculptures of the Cholas:

  1. Rajaraja and Rajendra built big temples of Thanjavur and Gangaikondacholapuram.
  2. They are architectural and sculptural marvels.
  3. Chola temples often became the centre of settlements around them.
  4. These were centres of craft production.
  5. Temples were also endowed with land by both rulers and others.
  6. The produce of this land went to maintain all the specialists who worked at the temple and very often lived near it.
  7. They included priests, garland makers, cooks, sweepers, musicians, dancers, etc.
  8. In other words, temples were not only places of worship; but also the hub of economic, social and cultural life.
  9. Amongst the crafts associated with temples, the making of bronze images was the most important.
  10. Chola bronze images are amongst the finest in the world.
  11. Most images were of deities, sometimes images were made of devotees also.

9. Look at Map 1 once more and find out whether there were any kingdoms in the state in which you live.
Answer:  The answer may vary as per the students’ state where they live. So the students should answer this question on the basis of their individual study of the Map.

10. Contrast the ‘elections’ in Uttaramerur with present day panchayat election?
Answer:  The system of election in Uttaramerur was quite different from that of the present day panchayat elections. In the Uttaramerur election, the names of the persons eligible to be members of the committees were written on small tickets of palm leaves. The tickets were put into an earthenware pot. Thereafter a young boy was asked to take out the tickets, one by one for each committee. But the present day panchayat members are elected through a general election.

The names of the total members are printed with a symbol allotted to them on a paper called ballot paper. The voter put a stamp over their choice and drops it in a ballot box. Votes are counted and the winners’ name is declared by the Returning Officer. Every procedure is organized by a statutory body.

11. Compare the temple shown in this chapter with any present-day temple in your neighbourhood, highlighting any similarities and differences that you notice.
Answer:  Attempt yourself.

12. Find out more about taxes that are collected at present. Are these in cash, kind, or labour services?
Answer:
Taxes collected at present.

  • Land tax
  • Income tax
  • Wealth tax
  • Sale tax
  • Property tax
  • Excise tax
  • VAT (value-added tax)
  • Service tax
  • Education cess {tax)
  • Tax on electricity
  • Water tax

These taxes are generally in cash

 Very Short Answer Type Questions

1. Who was Dantidurga?
Answer: Dantidurga was a Rashtrakuta chief who turned out to be a kshatriya after performing a ritual known as hiranya-garbha.

2. What was called ‘rent’?
Answer:  Resources obtained from the producers were called ‘rent’.

3. Who paid revenue?
Answer:  The traders paid revenue

4. What were ‘prashastis’?
Answer:  Prashastis were the details depicting the rule and personality of the rulers

5. Who were the writers of Prashastis?
Answer:  The learned brahmanas were the writers of the Prashastis.

6. Who was Kalhana?
Answer: Kalhana was a great Sanskrit poet from Kashmir.

7. What was ‘tripartite struggle’?             [V. Imp.]
Answer:  The tripartite sruggle was a long drawn conflict among the rulers of Gurjara- Pratihara, Rashtrakuta and Pala dynasties for control over Kanauj.

8. Who wrote Kitab al-Hind?
Answer:  A great Arab poet al-Baruni wrote Kitab al-Hind

9. Who was Prithviraja in ?
Answer: Prithviraja III was afamous Chauhan ruler of Delhi and surrounding regions during 1168-1192.

10. Who destroyed the temple of Somnath?
Answer: Sultan Mahmud of Ghazni, a ruler of Afghanistan, destroyed the temple of Somnath.

11 Why are the temples of Thanjavur and Gangaikonda-cholapuram famous for?[V. Imp.]
Answer:  These temples are famous for their architectural and sculptural marvels.

12. What was called as ‘ur’?
Answer:  The settlements of peaseants were called as ‘ur’.

13. What were called ‘nadu’?
Answer: The larger units of the groups of ‘ur’ were called ‘nadu’.

14. What was known as ‘brahmadeya’?
Answer: The land gifted to Brahmanas was called as ‘brahmadeya’.

15. Define ‘nagarams’.
Answer:  ‘Nagarams’ were the associationa of traders

16. What was ‘sabha’?
Answer:  An assembly of prominent Brahmanas that looked after brahmadeya was called ‘sabha’.

Short Answer Type Questions

1. Trace out the emergence of the new class of rulers.  [V. Imp.]
Answer: During the seventh century a new class of rulers emerged. They were the subordinates or samantas under the influence of big landlords or warriors. Later they themselves gained power and wealth and declared themselves to be maha- samantas, maha-mandaleshvara (the great lord of a circle or region)

2. Who was Dantidurga? How did he became a kshatriya?
Answer: Dantidurga was a Rashtrakuta chief in Deccan. Initially, Rashtrakutas were subordinate to the Chalukyas of Karnataka. Dantidurga, in the mid-eighth century, overthrew his Chalukya overlord and performed a ritual known as ‘hiranya-garbha’. It was thought to lead to the rebirth of the sacrificer as a kshatriya, even if he was not one by birth.

3. How was the financial position controlled in these states?    [V. Imp.]
Answer:  These states were closely associated with peasants, traders, and Brahmanas. The resources were obtained from peasants, cattle-keepers, and artisans who were the producers. They were compelled to surrender part of what they produced. The traders had to pay revenue. The functionaries for collecting revenue were generally from influential families or from the close relatives of the kings.

4. Write a note on ‘Prashastis’.
Answer:  Prashastis, often written by learned Brahmanas, were details of the functioning of a king. They may not be literally true. They tell us how rulers wanted to depict themselves—as valiant, victorious warriors etc. Brahmanas were rewarded by grants of land which were recorded on copper plates.

5. Who was Kalhana? Why was he famous for?
Answer:  Kalhana was a great Sanskrit poet who wrote a long Sanskrit poem depicting the ruler of Kashmir. He used a variety of sources, including inscriptions, documents, eyewitness accounts. He was different from the writers of Prashastis for he was often critical about rulers and their policies.

6. Why were the temples first targets of the conquerors?      [V. Imp.]
Answer:  It was the trend to demonstrate power and resources by building large temples. The rulers often invested a lot in the decoration and maintenance of the temples. So the foreign conquerors made them target first. The most prominent example of it was the temple 6f Somnath in Gujarat looted several times by Sultan Mahmud of Ghazani who came from Afghanistan. Much of the wealth he carried away was used to create a splendid capital city at Ghazani.

7. Who was Prithvirqja III? Who defeated him?
Answer:  Prithviraja III (1168-1192) was a Chahamanas, later known as Chauhan ruler. He ruled over the regions around Delhi and Ajmer. He defeated Sultan Muhammad Ghori, an Afghan conqueror in 1191. But only one year after it in 1192 he was defeated by Ghori. It was the end of his rule and life both.

8. What was the role of temples during Chola kingdom? [V. Imp.]
Answer:  The Chola rulers like Rajaraja and Rajendra built two great temples of Thanjavur and Ganaikonda-cholapuram respectively. Temples were the nuclei of settlements during this period. They were the centres of craft production and also controlled the financial aspects related to them. Hence they were not only the religious centres but also the hub of economic, social, and cultural life as well.

9 Describe the irrigation system of the Chola kingdom.    [Imp.]
Answer: The irrigation system during the Chola ruler was much improved. A variety of methods were applied to it. Many wells were dug; huge tanks were also constructed to collect water. The kings themselves took interest in organizing labour and resources and extensive planning.

10. What was the system of election of committee members for Sabha?
Answer:  The Sabha had separate committees to manage the works of irrigation and other works of gardens, temples, etc. The members of each committee were elected through a strict process. Names of eligible candidates were written on small tickets of palm leaf. The tickets were put into an earthenware pot. A young boy was asked to take out the tickets one by one for each committee.

Long Answer Type Questions

1. Write a note on Sultan Mahmud of Ghazni  [Imp.]
Answer:  Sultan Mahmud of Ghazni was a ruler of Afghanistan from 997 to 1030. He was an ambitious king and so made all efforts to extend his control over far off regions. He extended his control over parts of Central Asia, Iran, and the north­western part of the subcontinent. Almost every year he raided the subcontinent targeting wealthy temples. The temple of Somnath in Gujarat was the worst victim of his raids. Much of the wealth he carried away was used to create a splendid capital city at Ghazni. Ghazni was fond of knowing more about the people he conquered. He entrusted a scholar named al-Biruni to write an account of the subcontinent. The al- Biruni’s account, an Arabic work titled as Kitab al-Hind, is still an important source for historians.

2. How did the Cholas rise to power? Trace out the role of Rajaraja I in this rise.[V. Imp.]
Answer:  Cholas belonged to a minor chiefly family known as Muttaraiyar that held power in the Kaveri delta. They worked as subordinate to the Pallava kings of Kanchipuram. Vijayalaya, a member of the ancient chiefly family of the Cholas from Uraiyur, captured the delta from the Muttaraiyar in the middle of the ninth century. ‘Hie town of Thanjavur and the temple for goddess Nishmbhasudini are some of his great creations. Vijayalaya’s successors conquered neighboring regions and the kingdom grew in size and power.

The Pandyan and the Pallava territories to the south and north were made part of this kingdom.Rajaraja I was the most powerful Chola ruler. He became king in 985 and expanded control over most of these areas. He was well known for the reorganization of the administration of his empire. His son Rajendra I continued his policies an even raided the Ganga valley, Sri Lanka and contries of Southeast Asia, developing a navy for these expeditions. The big temples of Thanjavur and Gangaikonda-cholapuram built by them are architectural marvels.

Read More

RD SHARMA SOLUTION CHAPTER- 9 Trigonometric Ratios of Multiple and Submultiple Angles I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles Exercise Ex. 9.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Prove that

Solution 24

Question 26

Solution 26

Question 27

Solution 27

Question 28(i)

Solution 28(i)

Question 28(ii)

Solution 28(ii)

Question 29

Solution 29

Question 30(i)

Solution 30(i)

Question 30(ii)

Solution 30(ii)

Question 30(iii)

Solution 30(iii)

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 36

Solution 36

Question 37

Solution 37

Question 38(i)

Solution 38(i)

Question 38(ii)

Solution 38(ii)

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 25

Solution 25

Question 31

Solution 31

Question 35

Solution 35

Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles Exercise Ex. 9.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles Exercise Ex. 9.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Read More

Chapter – 1 Tracing Changes Through a Thousand Years | Class 7th | NCERT History Solutions | Edugrown

NCERT Solutions for Class 7 Social Science includes all the questions provided in NCERT Class 7 Social Science Text Book of Geography The Earth: Our Habitat, History Our Pasts, Civics Social and Political Life. Here CBSE Class 7 SST all questions are solved with the detailed explanation to score good marks in the exams. you can check Extra Questions for Class 7 Social Science

Chapter -1 Tracing Changes Through a Thousand Years

Question 1. Who was considered a ‘foreigner’ in the past?
Answer:  The term ‘foreigner’ is used in the sense of a person who is not an Indian. In the medieval period it was applied to any stranger who appeared, say in a given village, someone who was not a part of that society or culture. In this sense a forest-dweller was a foreigner for a city-dweller. But two peasants living in the same village were not foreigners to each other, even though they may have had different religious or caste backgrounds.

Question  2. State whether true or false:

  • We do not find inscriptions for the period after 700.
  • The Maraihas asserted their political importance during this period.
  • Forest-dwellers were sometimes pushed out of their lands with the spread of agricultural settlements.
  • Sultan Ghiyasuddin Balban controlled Assam, Manipur and Kashmir.

Answer:  (a) False; (b) False; (c) True; (d) False

Question 3. Fill in the blanks:
(a) Archives are places where………….. are kept.
(b) …………….was a fourteenth-century chronicler.
(c) ……., ……., ………, ……… and ………… were some of the crops introduced into the subcontinent during this period.
Answer:  (a) Manuscripts
(b) Ziyauddin Barani
(c) Potatoes, com, chillies, tea, coffee.

Question 4. List some of the technological changes associated with this period.
Answer:
Some of the technological changes associated with this period are:

  1. Persian wheel in irrigation.
  2. Spinning wheel.
  3. Fire-arms in combat.

Question 5. What were some of the major religious developments during this period?
Answer:  
Some of the major significant religious developments occurred in Hinduism. The worship of new deities, the construction of temples by royalty and growing importance of Brahmanas, the priests, as dominant groups in society were the new changes. Brahmanas’ importance grew due to their knowledge of Sanskrit language. They were patronized by the Emperors. The idea of bhakti emerged among people. The merchants and migrants brought with them the teachings of Quran, the holy book of Muslims.

Question 6. In what ways has the meaning of the term ‘Hindustan’ charged over the centuries?
Answer:
The meaning of the term ‘Hindustan’ has changed over the centuries in the following manner:

  1. In the thirteenth century Minhaj-i-Siraj used the term ‘Hindustan’. He meant areas of Punjab, Haryana and the lands between Ganga and Yamuna. He used this term in a political sense that were a part of the dominions of the Delhi Sultanate. The term never included South India.
  2. In the sixteenth century poet Babur used the term ‘Hindustan’ to describe the geography, the fauna and the culture of the inhabitants of the subcontinent.
  3. In fourteenth-century poet Amir Khusrau used the term ‘Hind’ in the same sense as Babur did in the sixteenth century.
  4. ‘Hindustan’ did not carry the political and national meanings as the term ‘India’ does today.

Question  7. How were the affairs of jatis regulated?
Answer:
The affairs of jatis were regulated in the following way:

  1. Jatis formed their own rules and regulations.
  2. There was an assembly of elders called jati panchayat.
  3. It enforced the rules and regulations.
  4. Jatis were also directed to follow the rules of the village.
  5. Several villages were governed by a chieftain.

Question 8. What does the term pan-regional empire mean?
Answer:  
The term ‘pan-regional’ was used in the sense of the areas of empires spanning diverge regions. The dynasties like Cholas, Khaljis, Tughluqs, and Mughals extended their empires pan-regional. Though, not all these empires were equally stable or successful. But pan-regional rule altered the character of the regions. Most of the regions across the subcontinent were left with the legacies of the big and small states that had ruled over them. The emergence of many distinct and shared traditions in governance the economy elite cultures and languages were some of the prominent factors that took place as a result of pan-regional rules.

Question 9. What are the difficulties historians face in using manuscripts?
Answer:
Historians faced a lot of difficulties while using manuscripts because:

  1. There was no printing press in the 13th and 14th centuries. Scribes in those days made manuscripts by hand.
  2. To copy was not an easy exercise. Scribes could not read the handwriting of the other writers.
  3. They were forced to guess. So there were small but significant differences in the copy of the scribed.
  4. These small words or sentences here and there grew over centuries of copying.
  5. The manuscripts of the same text became a great extent different from the original.

Question 10. How do historians divide the past into periods? Do they face any problems in doing so?
Answer:
Historians divide the past into periods on the basis of continuity. This continuity is further based on:

  1. Coins
  2. Inscriptions
  3. Architecture
  4. Textual records

But they face difficulties in doing so as discontinuity exists.

  1. Textual records increased tremendously.
  2. They gradually displaced other types of available information.

Thousand years of human history (or of any country or region) witnessed a number of changes. After all, the ancient history of India is different from that of the other two periods i.e., the medieval period and modem period. Therefore describing the entire period as one historical unit is not an easy task.

Question 11. Compare either Map 1 or Map 2 with the present-day map of the subcontinent listing as many similarities and differences as you can find.
Answer:  
Map 1 and Map 2 given in the NCERT Textbook represent two different times. Map 1 was made in 1154 CE by al-Idrisi, an Arab geographer. This section is a detail of the Indian subcontinent from his larger map of the world. Map 2 was made by a French cartographer in 1720. Both maps are quite different from each other, even though they represent the same area. In Map 1 we find south India at the place where we would expect to find north India and Sri Lanka is the island at the top. The place names are in Arabic. Some familiar places like Kanauj in Uttar Pradesh have been spelt as Qanauj. In comparison to this Map 2 was made nearly 600 later after Map 1. By that time information about the subcontinent had changed a lot. This map appears to be more familiar to us. The coastal areas, particularly, are more detailed.

Question 12. Find out where records are kept in your village or city. Who writes these records? Is there an archive? Who manages it? What kinds of documents are stored there? Who are the people who use it?
Answer:

  1. Records are kept in our city at the archives. These records are written by the officials of the Revenue Department.
  2. The in charge of the Archives/ Deputy Director of Archives manages these records.
  3. Rare manuscripts, government records, and other valuable books, etc. are stored there.
  4. Scholars, researchers, and government officials use them.

V. Very Short Answer Type Questions

Question 1. Who was al-Idrisi?
Answer:  al-Idrisi was an Arab cartographer.

Question 2. Who is a ‘cartographer’?
Answer:  Cartographer is one who draws a map.

Question 3. What difference do you notice in the map drawn by al-Idrisi?
Answer: In the map drawn by al-Idrisi we find a completely different view. Here south India is shown at present North India and Sri Lanka is the island at the top

Question 4. Who used the term Hindustan for the first time and when?
Answer:  Minhaj-i Siraj used the term ‘Hindustan’ for the first time in the thirteenth century.

Question 5. What sources do historians use for the study of a particular period of history?[V. Imp.]
Ans. The historians use sources like coins, inscriptions, architectures, and textual records for the study of a specific period.

Question 6. What do you mean by archives?
Answer:  Archives were the places where manuscripts were collected.

Question 7. Who were scribes?
Answer: Scribes were those professionals who used to copy down the manuscripts.

Question 8. How did the scribes copy down the manuscripts?
Answer: Scribes copied down the manuscripts by hand.

Question 9. What changes took place during 700 and 1750? [V. Imp.]
Answer: Many technologies like the Persian wheel in irrigation, the spinning wheel in weaving, and firearms in combat made their appearance. Some new foods and beverages like potatoes, corn, chilies, tea, and coffee also arrived in the subcontinent.

Question 10. What factors contributed to a variety of developments?
Answer: The new technologies and innovations came to the subcontinent with the people who came from other areas and settled here.

Question 11. What were the new groups of people to be prominent at this age? [V. Imp.]
Answer: Rajputs, Marathas, Sikhs, Jats, Ahoms, and Kayasthas were the groups which came to be prominent in this age. They availed most of the opportunities of society.

Question 12. What do you mean by Jati Panchayat?
Answer: Jati Panchayat was the assembly of elders that controlled the conduct of the members of their jati They had their own rules and regulations.

Question 13. Who was the Chief of the village?
Answer: Villages were controlled by a Chieftain. Even the smaller Jati Panchayats were bound to follow the village administration.

Question 14. What was the stretch of Delhi Sultan Ghiyasuddin Balban’s Empire?
Answer: According to a Sanskrit Prashasti Delhi Sultan Ghiyasuddin Balban’s empire was stretched from Bengal (Gauda) in the east to Ghazni (Gajjana) in Afghanistan in the west. It also included all of south India (Dravida)

Question 15. Why did Brahmanas dominate in society during this period? [Imp.]
Answer: Brahmanas were the only class of people who were proficient in the Sanskrit language. This was the reason that made them prominent.

Question 16. Who were the patrons?
Answer:  Patrons were a group of rulers and rich class of people who provided protection and livelihood to the Brahmanas, artists, and poets.

Question 17. What was the major development of this age?
Answer: The emergence of the idea of bhakti was the major development of this age.

Question 18. How history was divided by historians during the middle of the nineteenth century?
Answer: The British historians divided the history of India into three periods—:P

  • Hindu,
  • Muslim and
  • British.

Question 19. What was the basis of such division?
Answer: Such division was made on the basis of the religion as the historians did not consider any aspect more prominent other than the developments in religions.

Short Answer Type Questions

Question 1.  What difference do you trace out in the region of Hindustan of the thirteenth century and the modem India?
Answer: The term ‘Hindustan’ in the thirteenth century implied the areas of Punjab, Haryana, and the lands between the Ganga and Yamuna. Minhaj-i Siraj used the term in a political sense for lands consisting of a part of the dominions of the Delhi Sultanate. The areas included in this term shifted with the extent of the Sultanate. However, it never included south India.

Question 2. What does time mean for historians? How does it help them? [V. Imp.]
Answer:  Time, for historians, doesn’t mean just a passing of hours, days, or years. Instead, it reflects changes in social and economic organization, in the persistence and transformation of ideas and beliefs. In order to study historical developments historians divide the past into large segments. It makes the study convenient. The historians study different aspects of the specific period and then assess the comparative developments their impact on society and their contribution to the future generations.

Question 3. What do you mean by pan-regional rule? What was its impact? [V. Imp.]

Answer:  Pan-regional rule applies to the trend of extending the empire to the region beyond one’s own state. With the decline of the Mughal Empire in the eighteenth century, many regional states emerged. Consequently, a chance of sharing different traditions in the realms of governance, economy, elite cultures, and languages was brightened. People knew a lot of new things, manners, etc, without losing their own culture and identity.

Long Answer Type Questions

Question 1. What was the process of copying the manuscript? What were its drawbacks? [Imp.]
Answer: As there was no printing press during the period between 700 and 1750, Scribes used to copy down the manuscripts which were hand-written. Sometimes it was difficult to recognize the original script. So the Scribes used their own way of interpreting the facts. Consequently, there were differences were found in the copies written by different Scribes. As all the copies were handwritten, it was difficult to recognize which was the original one. It was the drawback of such copying.

Question 2. Trace out the major changes in society during 700 and 1750? What was its main reason? [V. Imp.]
Answer: A number of changes took place in society between 700 and 1750. This period traced the technological appearance of Persian wheel in irrigation, the spinning wheel in weaving and firearms in combat. Potatoes, com, chilies, tea, and coffee were some of the new foods and beverages. These developments came with the arrival of the people who migrated to this land. As a result this period saw various changes in the economic, political, social and cultural life,

Question 3. What was the change in the religion of the time? Trace out major developments? [V. Imp.]
Answer: The period between 700 and 1750 witnessed major changes in religion. It was seen prominently in Hinduism. The worship of new deities, the construction of temples by royalty, and the growing importance of Brahmanas, the priests, as dominant groups in society were some of the major developments.

The idea of bhakti emerged. Merchants and migrants brought the new teachings of the ‘Quran’, the holy book of the Muslims. A class of patrons emerged. They were the rulers who provided shelter and protection to the ulemas—the learned theologians and jurists. Muslims were divided into two groups—Shia and Sunni. Shia Muslims believed in Prophet Muhammad’s authority while the Sunnis accepted the authority of the early leaders—Khalifas.

Read More

RD SHARMA SOLUTION CHAPTER-8 Transformation Formulae ICLASS 11TH MATHEMATICS-EDUGROWN

Chapter 8 Transformation Formulae Exercise Ex. 8.1

Question 1

Solution 1

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 4

Solution 4

Question 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 5(iv)

Solution 5(iv)

Question 5(v)

Solution 5(v)

Question 5(vi)

prove that ten 20o tan 30o tan 40o tan 80o = 1Solution 5(vi)

Question 5(vii)

Solution 5(vii)

Question 5(viii)

Solution 5(viii)

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 7

Solution 7

Question 8

Solution 8

Chapter 8 Transformation Formulae Exercise Ex. 8.2

Question 3(i)

cos 55 to the power of 0 plus cos 65 to the power of 0 plus cos 175 to the power of 0 space equals space 0

Solution 3(i)

Question 18

I f space x cos theta equals y cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses equals z cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses comma space P r o v e space t h a t space x y plus y z plus z x equals 0

Solution 18

G i v e n space x cos theta equals y cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses equals z cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses equals k left parenthesis s a y right parenthesis
x equals fraction numerator k over denominator cos theta end fraction
y equals fraction numerator k over denominator cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses end fraction
z equals fraction numerator k over denominator cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses end fraction
x y plus y z plus z x equals k squared open square brackets fraction numerator 1 over denominator cos theta cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses end fraction plus fraction numerator 1 over denominator cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses end fraction plus fraction numerator 1 over denominator cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses cos theta end fraction close square brackets
equals k squared open square brackets fraction numerator cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses plus cos theta plus cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses over denominator cos theta cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses end fraction close square brackets
equals k squared open square brackets fraction numerator cos theta cos fraction numerator 4 straight pi over denominator 3 end fraction minus sin theta sin fraction numerator 4 straight pi over denominator 3 end fraction plus cos theta plus cos theta cos fraction numerator 2 straight pi over denominator 3 end fraction minus sin theta sin fraction numerator 2 straight pi over denominator 3 end fraction over denominator cos theta cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses end fraction close square brackets
equals k squared open square brackets fraction numerator cos theta open parentheses begin display style fraction numerator minus 1 over denominator 2 end fraction end style close parentheses minus sin theta open parentheses begin display style fraction numerator minus square root of 3 over denominator 2 end fraction end style close parentheses plus cos theta plus cos theta open parentheses begin display style fraction numerator minus 1 over denominator 2 end fraction end style close parentheses minus sin theta open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses over denominator cos theta cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses end fraction close square brackets
equals k squared open square brackets fraction numerator minus cos theta plus sin theta open parentheses begin display style fraction numerator square root of 3 over denominator 2 end fraction end style close parentheses plus cos theta plus minus sin theta open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses over denominator cos theta cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses end fraction close square brackets
equals 0

H e n c e space P r o v e d

Question 1

Solution 1

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Solution 2(v)

text LHS = end text sin 105 to the power of ring operator plus cos 105 to the power of ring operator
equals sin 105 to the power of ring operator plus cos left parenthesis 90 to the power of ring operator plus 15 to the power of ring operator right parenthesis
equals sin 105 to the power of ring operator minus sin 15 to the power of ring operator
equals 2 sin open parentheses fraction numerator 105 to the power of ring operator minus 15 to the power of ring operator over denominator 2 end fraction close parentheses cos open parentheses fraction numerator 105 to the power of ring operator plus 15 to the power of ring operator over denominator 2 end fraction close parentheses
equals 2 sin 45 to the power of ring operator cos 60 to the power of ring operator
equals 2 fraction numerator 1 over denominator square root of 2 end fraction 1 half
equals fraction numerator 1 over denominator square root of 2 end fraction
equals cos 45 to the power of ring operator

Question 2(vi)

Solution 2(vi)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

Question 3(vi)

Solution 3(vi)

Question 3(vii)

Solution 3(vii)

Question 3(viii)

Solution 3(viii)

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 6(iv)

Solution 6(iv)

Question 6(v)

Solution 6(v)

Question 6(vi)

Solution 6(vi)

Question 6(vii)

Prove that:

cosθcos straight theta over 2 minus cos 3 θcos fraction numerator 9 straight theta over denominator 2 end fraction equals sin 7 θsin 8 straight theta

Solution 6(vii)

table attributes columnalign left end attributes row cell Consider text   the   left   hand   side   of   the   given   expression : end text end cell row cell straight L. straight H. straight S. equals cosθcos straight theta over 2 minus cos 3 θcos fraction numerator 9 straight theta over denominator 2 end fraction end cell row cell We text   know   that   2 cosAcosB = cos end text left parenthesis text A + B end text right parenthesis plus cos left parenthesis straight A minus straight B right parenthesis end cell row cell Thus comma end cell row cell straight L. straight H. straight S. equals 1 half open square brackets text cos end text open parentheses straight theta text + end text straight theta over 2 close parentheses plus cos open parentheses straight theta minus straight theta over 2 close parentheses close square brackets minus 1 half open square brackets text cos end text open parentheses 3 straight theta text + end text fraction numerator 9 straight theta over denominator 2 end fraction close parentheses plus cos open parentheses 3 straight theta minus fraction numerator 9 straight theta over denominator 2 end fraction close parentheses close square brackets end cell row cell text           end text equals 1 half open square brackets text cos end text open parentheses fraction numerator 3 straight theta over denominator 2 end fraction close parentheses plus cos open parentheses straight theta over 2 close parentheses close square brackets minus 1 half open square brackets text cos end text open parentheses fraction numerator 15 straight theta over denominator 2 end fraction close parentheses plus cos open parentheses negative fraction numerator 3 straight theta over denominator 2 end fraction close parentheses close square brackets end cell row cell text           end text equals 1 half open square brackets text cos end text open parentheses fraction numerator 3 straight theta over denominator 2 end fraction close parentheses plus cos open parentheses straight theta over 2 close parentheses close square brackets minus 1 half open square brackets text cos end text open parentheses fraction numerator 15 straight theta over denominator 2 end fraction close parentheses plus cos open parentheses fraction numerator 3 straight theta over denominator 2 end fraction close parentheses close square brackets text   end text text              end text left square bracket because cos left parenthesis negative straight theta right parenthesis equals cosθ right square bracket end cell row cell text           end text equals 1 half open square brackets text cos end text open parentheses fraction numerator 3 straight theta over denominator 2 end fraction close parentheses plus cos open parentheses straight theta over 2 close parentheses minus text cos end text open parentheses fraction numerator 15 straight theta over denominator 2 end fraction close parentheses minus cos open parentheses fraction numerator 3 straight theta over denominator 2 end fraction close parentheses close square brackets end cell row cell text           end text equals 1 half open square brackets cos open parentheses straight theta over 2 close parentheses minus text cos end text open parentheses fraction numerator 15 straight theta over denominator 2 end fraction close parentheses close square brackets end cell row cell Also text   we   know   that , end text end cell row cell text cosD end text minus cosC equals 2 sin fraction numerator straight C plus straight D over denominator 2 end fraction sin fraction numerator straight C minus straight D over denominator 2 end fraction end cell row cell Therefore comma end cell row cell straight L. straight H. straight S. equals 1 half cross times 2 sin fraction numerator fraction numerator 15 straight theta over denominator 2 end fraction plus straight theta over 2 over denominator 2 end fraction sin fraction numerator fraction numerator 15 straight theta over denominator 2 end fraction minus straight theta over 2 over denominator 2 end fraction end cell row cell text         end text equals sin fraction numerator fraction numerator 16 straight theta over denominator 2 end fraction over denominator 2 end fraction sin fraction numerator fraction numerator 14 straight theta over denominator 2 end fraction over denominator 2 end fraction end cell row cell text        end text equals sin fraction numerator 8 straight theta over denominator 2 end fraction sin fraction numerator 7 straight theta over denominator 2 end fraction end cell row cell text        end text equals straight R. straight H. straight S. end cell row blank row cell asterisk times Note colon text   Question   given   in   the   book   is   incorrect. end text end cell row cell text R. H. S .  should   be   equal   to   end text sin fraction numerator 8 straight theta over denominator 2 end fraction sin fraction numerator 7 straight theta over denominator 2 end fraction. end cell end table

Question 7(i)

Solution 7(i)

Question 7(ii)

Solution 7(ii)

Question 7(iii)

Solution 7(iii)

Question 7(iv)

Solution 7(iv)

Question 7(v)

Solution 7(v)

Question 8(i)

Solution 8(i)

Question 8(ii)

Solution 8(ii)

Question 8(iii)

Solution 8(iii)

Question 8(iv)

Solution 8(iv)

Question 8(v)

Solution 8(v)

Question 8(vi)

Solution 8(vi)

Question 8(vii)

Solution 8(vii)

Question 8(viii)

Solution 8(viii)

text LHS end text equals fraction numerator s i n 3 A cos 4 A minus sin A cos 2 A over denominator sin 4 A sin A plus cos 6 A cos A end fraction
equals fraction numerator 2 left parenthesis s i n 3 A cos 4 A minus sin A cos 2 A right parenthesis over denominator 2 left parenthesis sin 4 A sin A plus cos 6 A cos A right parenthesis end fraction
equals fraction numerator 2 s i n 3 A cos 4 A minus 2 sin A cos 2 A over denominator 2 sin 4 A sin A plus 2 cos 6 A cos A end fraction
equals fraction numerator s i n left parenthesis 4 A plus 3 A right parenthesis minus s i n left parenthesis 4 A minus 3 A right parenthesis minus open square brackets s i n left parenthesis 2 A plus A right parenthesis minus s i n left parenthesis 2 A minus A right parenthesis close square brackets over denominator cos left parenthesis 4 A minus A right parenthesis minus cos left parenthesis 4 A plus A right parenthesis plus cos left parenthesis 6 A plus A right parenthesis plus cos left parenthesis 6 A minus A right parenthesis end fraction
equals fraction numerator s i n left parenthesis 7 A right parenthesis minus s i n left parenthesis A right parenthesis minus s i n left parenthesis 3 A right parenthesis plus s i n left parenthesis A right parenthesis over denominator cos left parenthesis 3 A right parenthesis minus cos left parenthesis 5 A right parenthesis plus cos left parenthesis 7 A right parenthesis plus cos left parenthesis 5 A right parenthesis end fraction
equals fraction numerator s i n left parenthesis 7 A right parenthesis minus s i n left parenthesis 3 A right parenthesis over denominator cos left parenthesis 3 A right parenthesis plus cos left parenthesis 7 A right parenthesis end fraction
equals fraction numerator 2 s i n open parentheses fraction numerator 7 A minus 3 A over denominator 2 end fraction close parentheses cos open parentheses fraction numerator 7 A plus 3 A over denominator 2 end fraction close parentheses over denominator 2 cos open parentheses fraction numerator 7 A plus 3 A over denominator 2 end fraction close parentheses cos open parentheses fraction numerator 7 A minus 3 A over denominator 2 end fraction close parentheses end fraction
equals fraction numerator s i n 2 A over denominator cos 2 A end fraction
equals tan 2 A
equals R H S

Question 8(ix)

Solution 8(ix)

Question 8(x)

Solution 8(x)

Question 8(xi)

Solution 8(xi)

Question 9(i)

Solution 9(i)

Question 9(ii)

Solution 9(ii)

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13(i)

Solution 13(i)

Question 13(ii)

Solution 13(ii)

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 19

If text   end text straight m text   end text sinθ equals text   end text straight n text   end text sin left parenthesis straight theta plus 2 straight a right parenthesis comma text   end text prove text   end text that text   end text tan left parenthesis straight theta plus straight a right parenthesis equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction tana

Solution 19

table attributes columnalign left end attributes row cell Given text   that   end text straight m text   end text sinθ equals text   end text straight n text   end text sin left parenthesis straight theta plus 2 straight a right parenthesis comma text   end text end cell row cell text We   need   to   end text prove text   end text that text    end text tan left parenthesis straight theta plus straight a right parenthesis equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction tana end cell row cell straight m text   end text sinθ equals text   end text straight n text   end text sin left parenthesis straight theta plus 2 straight a right parenthesis end cell row cell rightwards double arrow fraction numerator sin left parenthesis straight theta plus 2 straight a right parenthesis over denominator text   end text sinθ end fraction equals straight m over straight n end cell row cell text Using   Componendo-Dividendo ,  we   have , end text end cell row cell rightwards double arrow fraction numerator sin left parenthesis straight theta plus 2 straight a right parenthesis plus sinθ over denominator sin left parenthesis straight theta plus 2 straight a right parenthesis minus sinθ end fraction equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction.... left parenthesis 1 right parenthesis end cell row cell We text   know   that , end text end cell row cell text sinC + sinD = 2 sin end text fraction numerator text C + D end text over denominator text 2 end text end fraction text cos end text fraction numerator text C end text minus text D end text over denominator text 2 end text end fraction end cell row and row cell text sinC end text minus text sinD  =  2 cos end text fraction numerator text C + D end text over denominator text 2 end text end fraction text sin end text fraction numerator text C end text minus text D end text over denominator text 2 end text end fraction end cell row cell text Applying   the   above   formulae   in   equation  ( 1 ),  we   have , end text end cell row cell fraction numerator text 2 sin end text fraction numerator straight theta text + 2 a + end text straight theta over denominator text 2 end text end fraction text cos end text fraction numerator straight theta text + 2 a end text minus straight theta over denominator text 2 end text end fraction over denominator text 2 cos end text fraction numerator straight theta text + 2 a + end text straight theta over denominator text 2 end text end fraction text sin end text fraction numerator straight theta text + 2 a end text minus straight theta over denominator text 2 end text end fraction end fraction equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction end cell row cell rightwards double arrow fraction numerator text 2 sin end text left parenthesis straight theta plus straight a right parenthesis cosa over denominator text 2 cos end text left parenthesis straight theta plus straight a right parenthesis sina end fraction equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction end cell row cell rightwards double arrow fraction numerator text tan end text left parenthesis straight theta plus straight a right parenthesis over denominator tana end fraction equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction end cell row cell rightwards double arrow text tan end text left parenthesis straight theta plus straight a right parenthesis equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction cross times tana end cell row cell text Hence   proved. end text end cell end table

Read More

RD SHARMA SOLUTION CHAPTER- 7 Trigonometric Ratios of Compound Angles I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 5 Trigonometric Ratios of Compound Angles Exercise Ex. 7.1

Question 34

I f space tan space left parenthesis A plus B right parenthesis equals p comma space tan space left parenthesis A minus B right parenthesis equals q comma space t h e n space s h o w space t h a t space tan space 2 A equals fraction numerator p plus q over denominator 1 minus p q end fraction

Solution 34

R H S comma
fraction numerator p plus q over denominator 1 minus p q end fraction
equals fraction numerator tan left parenthesis A plus B right parenthesis plus tan thin space left parenthesis A minus B right parenthesis over denominator 1 minus tan left parenthesis A plus B right parenthesis. tan left parenthesis A minus B right parenthesis end fraction
equals fraction numerator begin display style fraction numerator tan space A plus tan thin space B over denominator 1 minus tan space A. tan space B end fraction end style plus begin display style fraction numerator tan space A minus tan space B over denominator 1 plus tan space A. tan space B end fraction end style over denominator 1 minus fraction numerator tan space A plus tan thin space B over denominator 1 minus tan space A. tan space B end fraction. fraction numerator tan space A minus tan space B over denominator 1 plus tan space A. tan space B end fraction end fraction
equals fraction numerator begin display style fraction numerator left parenthesis tan thin space A plus tan space B right parenthesis left parenthesis 1 plus tan space A. tan space B right parenthesis plus left parenthesis tan space A minus tan space B right parenthesis left parenthesis 1 minus tan space A. space tan thin space thin space B right parenthesis over denominator left parenthesis 1 minus tan space A. tan space B right parenthesis left parenthesis 1 plus tan space A. tan space B right parenthesis end fraction end style over denominator begin display style fraction numerator left parenthesis 1 minus tan space A. tan space B right parenthesis left parenthesis 1 plus tan space A. tan space B right parenthesis minus left parenthesis tan space A plus tan space B right parenthesis. left parenthesis tan space A minus tan space B right parenthesis over denominator left parenthesis 1 minus tan space A. space tan space B right parenthesis left parenthesis 1 plus tan space A. tan space B right parenthesis end fraction end style end fraction
equals fraction numerator tan space A plus tan space B plus tan squared A. tan space B plus tan space A. tan squared B plus tan space A minus tan thin space B minus tan thin space squared A. tan space B plus tan space A. tan squared B over denominator 1 minus tan squared space A. tan squared space B minus tan squared space A plus tan squared space B end fraction
equals fraction numerator 2 space tan space A plus 2 space tan space A. tan squared B over denominator left parenthesis 1 minus tan squared A right parenthesis left parenthesis 1 plus tan squared B right parenthesis end fraction equals fraction numerator 2 space tan space A left parenthesis 1 plus tan squared space B right parenthesis over denominator left parenthesis 1 minus tan squared space A right parenthesis left parenthesis 1 plus tan squared space B right parenthesis end fraction equals fraction numerator 2 space tan space A over denominator 1 minus tan squared space A end fraction equals tan space 2 A equals L H S
H e n c e space P r o v e d

Question 19

I f space fraction numerator sin space left parenthesis x plus y right parenthesis over denominator sin space left parenthesis x minus y right parenthesis end fraction equals fraction numerator a plus b over denominator a minus b end fraction comma space s h o w space t h a t space fraction numerator tan space x over denominator tan space y end fraction equals a over b

Solution 19

fraction numerator sin space x. cos space y plus sin space y. cos space x over denominator sin space x. cos space y minus sin space y. cos space x end fraction equals fraction numerator a plus b over denominator a minus b end fraction
rightwards double arrow fraction numerator sin space x. cos space y plus sin space y. cos space x plus sin space x. cos space y minus sin space y. cos space x over denominator sin space x. cos space y plus sin space y. cos space x minus sin space x. cos space y plus sin space y. cos space x end fraction equals fraction numerator a plus b plus a minus b over denominator a plus b minus a plus b end fraction left square bracket U sin g space C o m p o n e n d o space a n d space D i v i d e n d o right square bracket
rightwards double arrow fraction numerator 2 sin space x. cos space y over denominator 2 sin space y. cos space x end fraction equals fraction numerator 2 a over denominator 2 b end fraction
rightwards double arrow fraction numerator tan space x over denominator tan space y end fraction equals a over b
H e n c e space P r o v e d

Question 32

I f space a n g l e space theta space i s space d i v i d e d space i n t o space t w o space p a r t s space s u c h space t h a t space t h e space tan g e n t s space o f space o n e space p a r t space i s space lambda space p a r t
t i m e s space t h e space tan g e n t space o f space t h e space o t h e r comma space a n d space Ï• space i s space t h e i r space d i f f e r e n c e space t h e n space s h o w space t h a t space space
sin space theta equals fraction numerator lambda plus 1 over denominator lambda minus 1 end fraction sin space Ï•.

Solution 32

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Question 33

I f space tan space theta equals fraction numerator sin space alpha minus cos space alpha over denominator sin space alpha plus cos space alpha end fraction comma space t h e n space s h o w space t h a t space sin space alpha plus cos space alpha equals square root of 2 cos space theta

Solution 33

tan space theta equals fraction numerator sin space alpha minus cos space alpha over denominator sin space alpha plus cos space alpha end fraction
rightwards double arrow tan space theta equals space fraction numerator tan space alpha minus 1 over denominator tan space alpha plus 1 end fraction left square bracket D i v i d i n g space b o t h space N u m e r a t o r space a n d space D e n o m i n a t o r space b y space cos space alpha right square bracket
rightwards double arrow tan space theta equals fraction numerator tan space alpha minus tan space begin display style pi over 4 end style over denominator 1 plus tan space begin display style pi over 4 end style. tan space alpha end fraction
rightwards double arrow tan space theta equals tan space open parentheses alpha minus pi over 4 close parentheses
rightwards double arrow theta equals alpha minus pi over 4 space space space left square bracket R e m o v i n g space tan space f r o m space b o t h space s i d e s right square bracket
rightwards double arrow cos space theta equals cos open parentheses alpha minus pi over 4 close parentheses space left square bracket T a k i n g space cos space o n space b o t h space s i d e s right square bracket
rightwards double arrow cos space theta equals cos space alpha. cos space pi over 4 plus sin space alpha. sin space pi over 4
rightwards double arrow cos space theta equals cos space alpha. fraction numerator 1 over denominator square root of 2 end fraction plus sin space alpha. fraction numerator 1 over denominator square root of 2 end fraction
rightwards double arrow cos space theta equals fraction numerator cos space alpha plus sin space alpha over denominator square root of 2 end fraction
rightwards double arrow square root of 2 space cos space theta equals sin space alpha plus cos space alpha
H e n c e space P r o v e d

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 2(a)

Solution 2(a)

Question 2(b)

Solution 2(b)

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 7

Solution 7

Question 8(i)

Solution 8(i)

Question 8(ii)

Solution 8(ii)

Question 8(iii)

Solution 8(iii)

Question 9

Solution 9

Question 10

Solution 10

Question 11(i)

Solution 11(i)

Question 11(ii)

Solution 11(ii)

Question 11(iii)

Solution 11(iii)

Question 12(i)

Prove that:

Sin (60o – q) cos (30o + q) + cos (60o – q) sin (30o + q) = 1Solution 12(i)

Question 12(iii)

Prove that:

Solution 12(iii)

Question 12(ii)

Prove that:

Solution 12(ii)

Question 13

Solution 13

Question 14(i)

Solution 14(i)

Question 14(ii)

Solution 14(ii)

Question 15(i)

Solution 15(i)

Question 15(ii)

Solution 15(ii)

Question 16(i)

Solution 16(i)

Question 16(ii)

Solution 16(ii)

Question 16(iii)

Solution 16(iii)

Question 16(iv)

Solution 16(iv)

Question 16(v)

Solution 16(v)

Question 16(vi)

Solution 16(vi)

Question 17(i)

Solution 17(i)

Question 17(ii)

Solution 17(ii)

Question 17(iii)

Solution 17(iii)

Question 17(iv)

Solution 17(iv)

Question 18

Solution 18

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28(i)

Solution 28(i)

Question 28(ii)

Solution 28(ii)

Question 29(i)

Solution 29(i)

Question 29(ii)

Solution 29(ii)

Question 29(iii)

Solution 29(iii)

Question 30

Solution 30

Question 31

Solution 31

Chapter 17 Trigonometric Ratios of Compound Angles Exercise Ex. 7.2

Question 4

Solution 4

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 3

Solution 3

Read More

RD SHARMA SOLUTION CHAPTER-6 Graphs of Trigonometric Functions I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 6 Graphs of Trigonometric Functions Exercise Ex. 6.1

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 2(i)

Sketch the graph of the following pairs of functions on the same axes:

  1. y = sin x, y = sin begin mathsize 11px style open parentheses straight x plus straight pi over 4 close parentheses end style

Solution 2(i)

Question 2(ii)

Sketch the graph of the following pairs of functions on the same axes:

ii.  y = sin x, y = sin 3xSolution 2(ii)

Chapter 6 Graphs of Trigonometric Functions Exercise Ex. 6.2

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Chapter 6 Graphs of Trigonometric Functions Exercise Ex. 6.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Sketch the graph of the following functions on the same scale.

y = tan x, y = tan2 xSolution 9

Question 10

Solution 10

Read More

Chapter – 18 Wastewater Story | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter -18 Wastewater Story

Q.1. Fill in the blanks:
(a) Cleaning of water is process of removing __________
(b) Wastewater released by houses is called __________.
(c) Dried __________ is used as manure.
(d) Drains get blocked by __________ and __________.
Ans.(a) pollutants (b) sewage (c) sludge (d) (d) chemicals, kitchen waste

Q.2. What is sewage? Explain why it is harmful to discharge untreated sewage into rivers or seas.
Ans. Sewage is a liquid containing wastes which is disposed by households, industrial and agricultural activities in water. It is harmful to discharge untreated sewage into rivers or seas because it can pollute the whole sources of water. Sewage contains harmful substances and disease causing organisms. It is therefore dangerous to release untreated sewage in water.

Q.3. Why should oils and fats be not released in the drain? Explain.
Ans. Oils and fats should not be released in drains because they harden the soil in the pipes and block them. Fats get clogged in holes of the soil in the drain and block it. It does not allow the water to flow.

Q.4. Describe the steps involved in getting clarified water from wastewater.
Ans. Following steps are involved in the purification of water:


(i) Firstly all the physical impurities like stones, plastic bags, cans etc. are to be removed. It is done by passing the water through bar screens.
(ii) Then water is taken to grit and sand removal tank where impurities are removed by sedimentation.
(iii) Solid impurities and feaces etc. are collected from bottom of water. These impurities collected are called sludge.
(iv) Clarified water is cleaned of other impurities by aerator. All disease causing bacteria are removed by chlorination.

Q.5. What is sludge? Explain how it is treated.
Ans. Sludge is the collected solid waste from the wastewater during the treatment in water treatment plant. Sludge is decomposed in a separate tank by the anaerobic bacteria. Activated sludge is used as manure.

Q.6. Untreated human excreta is a health hazard. Explain.
Ans. Untreated human excreta can cause a lot of health related problems. It pollutes water, air and soil. The polluted water contain disease causing bacteria which can spread epidemics like cholera, meningitis etc.

Q.7. Name two chemicals used to disinfect water.
Ans. Chlorine and ozone

Q.8. Explain the Junction of bar screens in a wastewater treatment plant.
Ans. Bar screens clear the wastewater of all the physical impurities. Large size waste like napkins, plastics, cans etc. are removed from the wastewater through the bar screens.

Q.9. Explain the relationship between sanitation and disease.
Ans. Sanitation and disease are related to each other. Sanitation involves proper disposal of sewage and refuse from hou&e and public places. If sanitation is there, no disease will occur, but if sanitation is not there various types of disease will occur and spread. So sanitation should be kept to avoid disease.

Q.10. Outline your role as an active citizen in relation to sanitation.
Ans. As active citizen we should take care of our personal environmental sanitation. We should make people aware of the benefits of sanitation. We should help municipal corporations to cover all the open drains and remove disease causing substances thrown in open.

Q.11.Here is a crossword puzzle. Good luck!
NCERT Solutions for Class 7 Science Chapter 18 Wastewater Story Q11
Across:
3. Liquid waste products
4. Solid waste extracted in sewage treatment
6. A word related to hygiene
8. Waste matter discharged from human body
Down:
1. Used water
2. A pipe carrying sewage
5. Micro-organisms which cause cholera
7. A chemical to disinfect water

Ans
NCERT Solutions for Class 7 Science Chapter 18 Wastewater Story Q11.1

Q.12. Study the following statements about ozone:
(a) It is essential for breathing of living organisms.
(b) It is used to disinfect water.
(c) It absorbs ultraviolet rays.
(d) Its proportion in air is about 3%.
Which of these statements are correct?
(i) (a), (b) and (c) (ii) (b) and (c) (iii) (a) and (d) (iv) All four
Ans. (ii) (b) and (c)

Read More

RD SHARMA SOLUTION CHAPTER- 5 Trigonometric Functions I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 5 Trigonometric Functions Exercise Ex. 5.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Prove that cosθ(tanθ+2)(2 tanθ+1)=2 secθ+5 sinθSolution 16

Question 17

I f space x equals fraction numerator 2 space sin space theta over denominator 1 plus cos space plus sin space theta end fraction comma space t h e n space p r o v e space t h a t space fraction numerator 1 minus cos space plus sin space theta over denominator 1 plus sin space theta end fraction i s space a l s o space e q u a l space t o space x.

Solution 17

fraction numerator 2 space sin space theta over denominator 1 plus cos space theta plus sin space theta end fraction equals x
rightwards double arrow fraction numerator 2 space sin space theta left parenthesis 1 minus cos space theta plus sin space theta right parenthesis over denominator left parenthesis 1 plus cos space theta plus sin space theta right parenthesis left parenthesis 1 minus cos space theta plus sin space theta right parenthesis end fraction equals x space space space left square bracket R a t i o n a l i z i n g space t h e space d e n o m i n a t o r right square bracket
rightwards double arrow fraction numerator 2 space sin space theta left parenthesis 1 minus cos space theta plus sin space theta right parenthesis over denominator left parenthesis 1 plus sin space theta right parenthesis squared minus cos squared space theta end fraction equals x
rightwards double arrow fraction numerator 2 space sin space theta minus 2 space sin space theta cos space theta plus 2 space sin squared theta over denominator 1 plus sin squared space theta plus 2 sin space theta minus cos squared space theta end fraction equals x
rightwards double arrow fraction numerator 2 space sin space theta left parenthesis 1 plus cos space theta minus sin space theta right parenthesis over denominator 2 sin squared space theta plus 2 space sin space theta end fraction equals x
rightwards double arrow fraction numerator 2 space sin space theta left parenthesis 1 plus cos space theta minus sin space theta right parenthesis over denominator 2 sin space theta left parenthesis 1 plus sin space theta right parenthesis end fraction equals x
rightwards double arrow fraction numerator 1 plus cos space theta space minus sin space theta over denominator 1 plus sin space theta end fraction equals x space left square bracket C a n c e l l i n g space t h e space 2 space sin space theta space i n space b o t h space N u m e r a t o r space a n d space D e n o m i n a t o r right square bracket
H e n c e space P r o v e d

Question 18

Solution 18

Question 19

I f space tan space theta equals a over b comma space t h e n space f i n d space t h e space v a l u e space o f space square root of fraction numerator a plus b over denominator a minus b end fraction end root plus square root of fraction numerator a minus b over denominator a plus b end fraction end root.

Solution 19

square root of fraction numerator a plus b over denominator a minus b end fraction end root plus square root of fraction numerator a minus b over denominator a plus b end fraction end root
equals square root of fraction numerator begin display style a over b end style plus 1 over denominator begin display style a over b end style minus 1 end fraction end root plus square root of fraction numerator begin display style a over b end style minus 1 over denominator begin display style a over b end style plus 1 end fraction end root space left square bracket D i v i d i n g space b o t h space N u m e r a t o r space a n d space d e n o m i n a t o r space b y space b right square bracket
equals square root of fraction numerator tan space theta plus 1 over denominator tan space theta minus 1 end fraction end root plus square root of fraction numerator tan space theta minus 1 over denominator tan space theta plus 1 end fraction end root
equals square root of fraction numerator begin display style fraction numerator sin space theta over denominator cos space theta end fraction end style plus 1 over denominator begin display style fraction numerator sin space theta over denominator cos space theta end fraction end style minus 1 end fraction end root plus square root of fraction numerator begin display style fraction numerator sin space theta over denominator cos space theta end fraction end style minus 1 over denominator begin display style fraction numerator sin space theta over denominator cos space theta end fraction end style plus 1 end fraction end root
equals square root of fraction numerator begin display style fraction numerator sin space theta plus cos space theta over denominator cos space theta end fraction end style over denominator begin display style fraction numerator sin space theta minus cos space theta over denominator cos space theta end fraction end style end fraction end root plus square root of fraction numerator begin display style fraction numerator sin space theta minus cos space theta over denominator cos space theta end fraction end style over denominator begin display style fraction numerator sin space theta plus cos space theta over denominator cos space theta end fraction end style end fraction end root
equals square root of fraction numerator sin space theta plus cos space theta over denominator sin space theta minus cos space theta end fraction end root plus square root of fraction numerator sin space theta minus cos space theta over denominator sin space theta plus cos space theta end fraction end root
equals fraction numerator sin space theta plus cos space theta plus sin space theta minus cos space theta over denominator square root of sin squared space theta minus cos squared space theta end root end fraction
equals fraction numerator 2 space sin space theta space over denominator square root of sin squared space theta space minus space cos squared space theta end root end fraction

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Syntax error from line 1 column 702 to line 1 column 707. Unexpected '</mi>'.

Solution 24

L H S space equals space a b plus a minus b plus 1
equals left parenthesis s e c space theta minus tan space theta right parenthesis left parenthesis cos e c space plus c o t space right parenthesis plus s e c space theta minus tan space theta minus cos e c space theta minus c o t space theta plus 1
equals open parentheses fraction numerator 1 over denominator cos space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction close parentheses open parentheses fraction numerator 1 over denominator sin space theta end fraction plus fraction numerator cos space theta over denominator sin space theta end fraction close parentheses plus fraction numerator 1 over denominator cos space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction minus fraction numerator 1 over denominator sin space theta end fraction minus fraction numerator cos space theta over denominator sin space theta end fraction plus 1
equals fraction numerator 1 over denominator sin space theta space cos space theta end fraction plus fraction numerator 1 over denominator cos space theta end fraction cross times fraction numerator cos space theta over denominator sin space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction cross times fraction numerator 1 over denominator sin space theta end fraction minus tan space theta cross times c o t space theta plus fraction numerator 1 over denominator cos space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction minus fraction numerator 1 over denominator sin space theta end fraction minus fraction numerator cos space theta over denominator sin space theta end fraction plus 1
equals fraction numerator 1 over denominator sin space theta space cos space theta end fraction plus fraction numerator 1 over denominator sin space theta end fraction minus fraction numerator 1 over denominator cos space theta end fraction minus 1 plus fraction numerator 1 over denominator cos space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction minus fraction numerator 1 over denominator sin space theta end fraction minus fraction numerator cos space theta over denominator sin space theta end fraction plus 1
equals fraction numerator 1 over denominator sin space theta space cos space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction minus fraction numerator cos space theta over denominator sin space theta end fraction
equals fraction numerator 1 over denominator sin space theta space cos space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction minus fraction numerator cos space theta over denominator sin space theta end fraction
equals fraction numerator 1 minus sin squared space theta minus cos squared space theta over denominator sin space theta. cos space theta end fraction
equals fraction numerator 1 minus left parenthesis cos squared space theta plus sin squared space theta right parenthesis over denominator sin space theta. space cos space theta end fraction
equals fraction numerator 1 minus 1 over denominator sin space theta. cos space theta end fraction equals 0 equals R H S. space H e n c e space P r o v e d

Question 25

Solution 25

Question 26(i)

Solution 26(i)

Question 26(ii)

Solution 26(ii)

Question 26(iii)

Solution 26(iii)

Chapter 5 Trigonometric Functions Exercise Ex. 5.2

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Chapter 5 Trigonometric Functions Exercise Ex. 5.3

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 1(ix)

Solution 1(ix)

Question 1(x)

Solution 1(x)

Question 1(xi)

Solution 1(xi)

Question 1(xii)

Solution 1(xii)

Question 1(xiii)

Solution 1(xiii)

Question 1(xiv)

Solution 1(xiv)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Solution 2(v)

Question 2(vi)

Solution 2(vi)

Question 2(vii)

Solution 2(vii)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

Question 4

Solution 4

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 7

Solution 7

Question 8(i)

Solution 8(i)

Question 8(ii)

Solution 8(ii)

Question 9(i)

Solution 9(i)

Question 9(ii)

Solution 9(ii)

Question 9(iii)

Solution 9(iii)

Question 9(iv)

Solution 9(iv)

Question 9(v)

Solution 9(v)

Read More

Chapter – 17 Forests Our Lifeline | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter -17 Forests Our Lifeline

Q.1. Explain how animals dwelling in the forest help it grow and regenerate.
Ans. Animals help in growing and regenerating forests in many ways. Animals work as the cleaning agents in the forest. Microorganisms work on dead bodies of plants and animals and degenerate them. An’imals also help in pollination which helps in growing a number of plants. Herbivores helps the carnivores to grow as they serve as food for them. Thus flora and fauna mutually grow in the forest.

Q.2. Explain how forests prevent floods.
Ans. Forests can absorb a lot of water. The roots of the trees absorb the water and prevent it from flowing away. Roots of trees also help in percolation of water into the soil. This helps in preventing floods.

Q.3. What are decomposers? Name any two of them. What do they do in the forest?
Ans. Decomposers are the organisms which feed on the dead bodies of plants and animals. They clean the forests decaying dead bodies and replenishing the nutrients back to the forest soil, e.g.„ beetles and grubs.

Q.4. Explain the role of forest in maintaining the balance between oxygen and carbon dioxide in the atmosphere.
Ans. Plants release oxygen in the atmosphere during the process of photosynthesis. This oxygen is inhaled by the animals for respiration. During respiration, they release carbon-dioxide which is absorbed by plants. In this way the oxygen and carbon dioxide cycle goes on. Since forests contain a large number of plants, they help much in this cycle and maintain balance in nature.

Q.5. Explain why there is no waste in a forest.
Ans. There is no waste in the forest because decomposers convert all the dead bodies of the plants and animals into the humus which gets added to the soil. Thus, no waste remains.

Q.6. List five products we get from forests.
Ans. (i) We get medicines from forests.
(ii) We get gum from forests.
(iii) We get wood which is used for many purposes like making furniture, paper etc.
(iv) We get food for animals from forests.
(v) We get sealing wax from forests.

Q.7.Fill in the blank:
(a) The insects, butterflies, honeybees and birds help flowering plants in .
(b) A forest is a purifier of and .
(c) Herbs form the layer in the forest.
(d) The decaying leaves and animal droppings in a forest enrich the .
Ans. (a) pollination (b) water, air (c) lowest (d) soil as humus.

Q.8. Why should we worry about the conditions and issues related to forests far from us?
Ans. We should be worried about deforestation as it would lead to floods, increase in earth’s temperature, decreasing animals habitats and soil erosion. Damage to forests directly or indirectly affects human habitat and environment so it must be a matter of concern among us.

Q.9. Explain why there is a need of variety of animals and plants in a forest.
Ans. All plants and animals sustain the forest life and also C02 – 02 cycle goes on due to animals and plants. Animals convert the dead and decaying matters into humus and increase the fertility of soil, thus enhancing plant growth. All food chains and food webs need variety of plants and animals.

Q.10. In fig. 1 7.15 the artist has forgotten to put the labels and directions on the arrows. Mark the directions on the arrows and label the diagram using the following labels: clouds, rain, atmosphere, cabon dioxide, oxygen, plants, animals, soil, roots, water table.
NCERT Solutions for Class 7 Science Chapter 17 Forests Our Lifeline Q10

Q.11.Which of the following is not a forest product?
(i) Gum (ii) Plywood (iii) Sealing wax (iv) Kerosene
Ans.(iv) Kerosene

Q.12.Which of the following statements is not correct?
(i) Forests protect the soil from erosion.
(ii) Plants and animals in a forest are not dependent on one another.
(iii) Forests influence the climate and water cycle.
(iu) Soil helps forests to grow and regenerate.
Ans.(ii) Plants and animals in a forest are not dependent on one another.

Q.13.Micro-organisms act upon the dead plants to produce
(i) sand (ii) mushrooms (iii) humus (iu) wood
Ans. (i) Humus

Read More

RD SHARMA SOLUTION CHAPTER- 4 Measurement of Angles I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 4 Measurement of Angles Exercise Ex. 4.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Read More