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

Page / Exercise

Ex. 10
MCQ

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 = 90^o and BAC =

Solution 2

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

By Pythagoras theorem, we have

Solution 3

Solution 4

Solution 5

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

Solution 6

Solution 10

Solution 15

Given:

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

By Pythagoras theorem, we have

Solution 16

Given:

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

By Pythagoras theorem, we have

Solution 17

Given:

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

⇒ tan²θ + cot² θ + 2 tan θ cot θ = 25

⇒ tan²θ + cot² θ + 2 tan θ ×

⇒ tan²θ + cot² θ + 2 = 25

⇒ tan²θ + cot² θ = 23

Correct Option: D

Solution 9

We have,

(cosθ + sec θ) =

⇒ (cosθ + sec θ)² =

⇒ cos²θ + sec² θ + 2 cos θ sec θ =

⇒ tan²θ + cot² θ + 2 cos θ ×

⇒ tan²θ + cot² θ + 2 =

⇒ tan²θ + 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

Other Chapters for CBSE Class 10 Maths

Chapter 1- Real Numbers Chapter 2- Polynomials Chapter 3- Linear equations in two variables Chapter 4- Quadratic Equations Chapter 5- Arithmetic Progressions Chapter 6- Co-ordinate Geometry Chapter 7- Triangles Chapter 8- Circles Chapter 9- Constructions Chapter 11- T-Ratios of Some Particular Angles Chapter 12- Trigonometric Ratios of Complementary Angles Chapter 13- Trigonometric Identities Chapter 14- Height and Distance Chapter 15- Perimeter and Areas of Plane Figures Chapter 16- Areas of Circle, Sector and Segment Chapter 17- Volume and Surface Areas of Solids Chapter 18- Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Chapter 19- Probability

R S AGGARWAL AND V AGGARWAL Solutions for CBSE Class 10 Subjects