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- Introduction to Trigonometry : Important Formula (Trigonometry class ten) and Chapter Summary

In the right angle triangle ABC, the right angle is ∠ B.

The value of sin or cos never exceeds 1, but the value of sec and cosec is always greater than or equal to 1.

Exercise 8.1

Question 1 - In triangle ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:

(a) Sin A, Cos A

Solution: AB = 24 cm, BC = 7 cm, AC = ?

The value of AC can be calculated by using Pythagoras Theorem:

(b) Sin C, Cos C

Solution:

Question 2 - In the given figure, find tan P cot R.

Solution: Value of QR can be calculated by using Pythagoras theorem:

Now;

Question 3 - If sin A = ¾, calculate cos A and tan A.

Solution: Sin A = ¾ = p/h

We can calculate b by using Pythagoras theorem;

Now;

Question 4 - Given 15 cot A = 8, find sin A and sec A.

Solution: 15 cot A = 8

This means, b = 8 and p = 15

We can calculate h by using Pythagoras theorem;

Now;

Question 5 - Given sec Î¸ = 13/12, calculate all other trigonometric ratios.

Solution:

This means, h = 13 and b = 12.

We can calculate p by using Pythagoras theorem;

Other trigonometric ratios can be calculated as follows:

Question 6 - If ∠ A and ∠ B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B

Solution: For no two different angles the cos ratio is same. (Ref: Table of trigonometric ratio).

Question 7 - If cot Î¸ = 7/8, evaluate:

Solution:

This means, b = 7 and p = 8.

We can calculate h by using Pythagoras theorem;

Solution:

Solution:

Question 8 - If 3 cot A = 4, check whether or not.

Solution: 3 cot A = 4 means cot A = 4/3 = b/p

Hence, p = 3 and b = 4.

We can calculate h by using Pythagoras theorem;

Now; the equation can be checked as follows:

LHS:

RHS:

It is clear that LHS = RHS.

Question 9 - In triangle ABC, right-angled at B, if tan A = find the value of:

Solution:

We can calculate h by using Pythagoras theorem;

(a) Sin A Cos C + Cos A Sin C

Solution:

Sin A Cos C + Cos A Sin C

(b) Cos A Cos C Sin A Sin C

Solution:

Cos A Cos C Sin A Sin C

Question 10 - In triangle PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Solution: Given; PR + QR = 25 cm and PQ = 5 cm.

Hence, PR = 25 QR

We can calculate PR and QR by using Pythagoras theorem;

Now;

Question 11 - State whether the following are true or false. Justify your answer.

(a) The value of tan A is always less than 1.

Solution: False; value of tan begins from zero and goes on to become more than 1.

(b) sec A = 12/5 for some value of angle A.

Solution: True, value of cos is always more than 1.

(c) cos A is the abbreviation used for the cosecant of angle A.

Solution: False, cos is the abbreviation of cosine.

(d) cot A is the product of cot and A.

Solution: False, cot A means cotangent of angle A.

(e) sin A = 4/3 for some angle A.

Solution: False, value of sin is less than or equal to 1, while this value is more than 1.

- 10th Maths Paper Solutions Set 2 : CBSE Delhi Previous Year 2015
- 10th Maths Paper Solutions Set 3 : CBSE Delhi Previous Year 2015
- 10th Maths Paper Solutions Set 1 : CBSE Delhi Previous Year 2015
- 10th Maths Paper Solutions Set 1 : CBSE Abroad Previous Year 2015
- 10th Maths Paper Solutions Set 1 : CBSE All India Previous Year 2015

Introcution To Trigonometry will be available online in PDF book form soon. The solutions are absolutely Free. Soon you will be able to download the solutions.

Introduction
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NCERT Solution - Exercise 3.1
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NCERT Solution - Exercise 3.2 Part-1
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NCERT Solution - Exercise 3.2 Part-2
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NCERT Solution - Exercise 3.3 Part-1
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NCERT Solution - Exercise 3.3 Part-2
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NCERT Solution - Exercise 3.4 Part-1
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NCERT Solution - Exercise 3.4 Part-2
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NCERT Solution - Exercise 3.5 Part-1
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NCERT Solution - Exercise 3.5 Part-2
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NCERT Solution - Exercise 3.6 Part-1
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NCERT Solution - Exercise 3.6 Part-2
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NCERT Solution - Exercise 4.1 Part-1
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NCERT Solution - Exercise 4.1 Part-2
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NCERT Solution - Exercise 4.2 Part-1
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NCERT Solution - Exercise 4.2 Part-2
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NCERT Solution - Exercise 4.2 Part-3
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NCERT Solution - Exercise 4.3 Part-1
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NCERT Solution - Exercise 4.3 Part-2
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NCERT Solution - Exercise 4.3 Part-3
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NCERT Solution - Exercise 4.3 Part-4
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NCERT Solution - Exercise 4.3 Part-5
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NCERT Solution - Exercise 4.4 Part-1
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NCERT Solution - Exercise 4.4 Part-2
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Solution of Exercise 5.1 (NCERT)-Part-1
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Solution of Exercise 5.1 (NCERT)-Part-2
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Solution of Exercise 5.1 (NCERT)-Part-3.1
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Solution of Exercise 5.1 (NCERT)-Part-3.2
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Solution of Exercise 5.2 (NCERT) Part -1
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Solution of Exercise 5.2 (NCERT) Part -2
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Solution of Exercise 5.2 (NCERT) Part -3
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Solution of Exercise 5.2 (NCERT) Part -4
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Solution of Exercise 5.2 (NCERT) Part -5
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Solution of Exercise 5.2 (NCERT) Part -6
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Solution of Exercise 5.3 (NCERT)Part -1
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Solution of Exercise 5.3 (NCERT)Part - 2
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Solution of Exercise 5.3 (NCERT)Part - 3
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Solution of Exercise 5.3 (NCERT)Part - 4
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Solution of Exercise 5.3 (NCERT)Part - 5
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Solution of Exercise 5.3 (NCERT)Part - 6
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Solution of Exercise 5.3 (NCERT)Part - 7
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Solution of Exercise 5.3 (NCERT)Part - 8
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Solution of NCERT Exercise 5.4 (Optional)
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Introduction : Theorems
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NCERT Solution of Exercise 6.1
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NCERT Solution Exercise 6.2 Part - I
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NCERT Solution Exercise 6.2 Part - II
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NCERT Solution Exercise 6.2 Part - III
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NCERT Solution of Exercise 6.3 Part - III
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NCERT Solution of Exercise 6.4
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NCERT Solution of Exercise 6.5
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NCERT Solution of Exercise 6.5 Part - 2
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NCERT Solution of Exercise 6.6 - Part - 1
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NCERT Solution of Exercise 6.6 - Part - 2
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THEOREM 1:
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Solution of NCERT Exercise 10.1
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Solution of NCERT Exercise 10.2
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Solution of NCERT Exercise 10.2 - Part 2
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Solution of NCERT Exercise 10.2 - part - 3
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Solution of NCERT Exercise 10.2 - part - 4
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Solution of NCERT Exercise 10.2 - part - 5
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Solution of NCERT Exercise 10.2 - part - 6
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