# R S AGGARWAL AND V AGGARWAL Solutions for Class 10 Maths Chapter 10 - Trigonometric Ratios

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## Chapter 10 - Trigonometric Ratios Ex. 10

Solution 7

…… given

In right ∆ABC, ∠B = 90° and ∠A = θ

Let BC = c and AC =

Then, AB2 = AC2 - BC2 = c2 + d2 - c2 = d2

Solution 8

In right ∆ABC, ∠B = 90° and ∠A = θ

√3 tan θ = 1 ⇒ tan θ =

Let BC = k and AB = √3k

Then, AC2 = AB2 + BC2 = 3k2 + k2 = 4k2

Solution 9

In right ∆ABC, ∠B = 90° and ∠A = θ

4 tan θ = 3 ⇒ tan θ

Let BC = 3k and AB = 4k

Then, AC2 = AB2 + BC2 = 16k2 + 9k2 = 25k2

Solution 11

In right ∆ABC, ∠B = 90° and ∠A = θ

AC2 = AB2 + BC2 = b2 + a2

Solution 12

Solution 13

In right ∆ABC, ∠B = 90° and ∠A = θ

Let BC = k and AB = 2k

Then, AC2 = AB2 + BC2 = (4k2 + k2)= 5k2

Solution 14

Solution 18

Solution 19

Let BC = 3k and AC = 4k

Then, AB2 = AC2 - BC2 = (16k2 - 9k2)= 7k2

Solution 20(i)

In right ∆ABC, ∠B = 90° and ∠A = θ

Let BC = 4k and AB = 3k

Then, AC2 = AB2 + BC2 = (9k2 + 16k2)= 25k2

Solution 20(ii)

In right ∆ABC, ∠B = 90° and ∠A = θ

Let BC = 4k and AB = 3k

Then, AC2 = AB2 + BC2 = (9k2 + 16k2)= 25k2

Solution 21

Solution 22

In right ∆ABC, ∠B = 90° and tan A = 1

Let BC = k and AB = k

Then, AC2 = AB2 + BC2 = (k2 + k2)= 2k2

Solution 23

In right ∆ABC, ∠B = 90°

Then, BC2 = AC2 - AB2 = (17)2 - (8)2= 289 - 64 = 225

⇒ BC = √225 = 15cm

Therefore Length = 15cm and Breadth = 8cm

(i) The area of rect. ABCD = Length × Breadth = 15 × 8 = 120cm2

(ii) The perimeter of rect. ABCD =2(l + b) = 2(15 + 8) = 46cm

Solution 27

Solution 28

Solution 1

Given:

Let us draw a ABC in which B = 90o and BAC =

Solution 2

Let us draw a ABC in which B = 90o and BAC =

By Pythagoras theorem, we have

Solution 3

Solution 4

Solution 5

Let us draw a ABC in which B = 90o and BAC =

Solution 6

Solution 10

Solution 15

Given:

Let us draw a triangle ABC in which B = 90o and A =

By Pythagoras theorem, we have

Solution 16

Given:

Let us draw a triangle ABC in which B = 90o and A =

By Pythagoras theorem, we have

Solution 17

Given:

Let us draw a triangle ABC in which B = 90o and A =

By Pythagoras theorem, we have

Solution 24

Solution 25

Solution 26

Solution 29

Consider two right triangles XAY and WBZ such that sin A = sin B

Solution 30

Consider two right triangles XAY and WBZ such that tan A = tan B

## Chapter 10 - Trigonometric Ratios MCQ

Solution 1

Let BC = 8k and AB = 15k

Then, AC2 = AB2 + BC2 = (225k2 + 64k2) = 289k2

AC2 =289k2

Correct Option: C

Solution 2

tan θ

Let BC = √3k and AB = k

Then, AC2 = AB2 + BC2 = (k2 + 3k2) = 4k2

AC2 =4k2

Correct Option: A

Solution 3

cosec θ = √10

Let BC = k and AC = √10k

Then, AB2 = AC2 - BC2 = (10k2 - k2) = 9k2

AB2 =9k2

⇒ AB = √9k2 = 3k

Correct Option: D

Solution 4

Correct Option: B

Solution 5

Let AB = 4k and AC = 5k

Then, BC2 = AC2 - AB2 = (25k2 - 16k2) = 9k2

⇒ BC = √9k2 = 3k

Correct Option: A

Solution 6

Let AB = 4k and AC = 5k

Then, BC2 = AC2 - AB2 = (25k2 - 16k2) = 9k2

⇒ BC = √9k2 = 3k

Correct Option: A

Solution 7

Correct Option: D

Solution 8

We have

(tan θ + cot θ) = 5

⇒ (tan θ + cot θ)2 = 52

⇒ tan2 θ + cot2 θ + 2 tan θ cot θ = 25

⇒ tan2 θ + cot2 θ + 2 tan θ ×

⇒ tan2 θ + cot2 θ + 2 = 25

⇒ tan2 θ + cot2 θ = 23

Correct Option: D

Solution 9

We have,

(cosθ + sec θ) =

⇒ (cosθ + sec θ)2 =

⇒ cos2 θ + sec2 θ + 2 cos θ sec θ =

⇒ tan2 θ + cot2 θ + 2 cos θ ×

⇒ tan2 θ + cot2 θ + 2 =

⇒ tan2 θ + cot2 θ =

Correct Option: A

Solution 10

Let BC = 3k and AB = 4k

Then, AC2 = AB2 + BC2 = (4k2 + 3k2) = 25k2

Correct Option: B

Solution 11

Correct Option: C

Solution 12

Let AB = 2k and AC = 3k

Then, BC2 = AC2 - AB2 = (9k2 - 4k2) = 5k2

Correct Option: A

Solution 13

We have, sec θ + tan θ + 1 = 0 ⇒ sec θ + tan θ = -1

We know that,

(sec2 θ - tan2 θ) = 1

⇒ (sec θ - tan θ)(sec θ + tan θ) = 1

⇒ (sec θ - tan θ) × -1 =1

⇒ (sec θ - tan θ) = -1

Correct Option: B

Solution 14

cos A + cos2 A = 1….given

⇒ cos A = (1 - cos2 A) = sin2 A

∴ (sin2 A + sin4 A) = (cos A + cos2 A) = 1

Correct Option: C

Solution 15

Correct Option: D

Solution 16

Correct Option: A

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