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

## Chapter 10 - Trigonometric Ratios Ex. 10

…… given

In right ∆ABC, ∠B = 90° and ∠A = θ

Let BC = c and AC =

Then, AB^{2 }=
AC^{2} - BC^{2} = c^{2} + d^{2} - c^{2}
= d^{2}

In right ∆ABC, ∠B = 90° and ∠A = θ

√3 tan θ = 1 ⇒ tan θ =

Let BC = k and AB = √3k

Then, AC^{2 }=
AB^{2} + BC^{2} = 3k^{2} + k^{2} = 4k^{2}

In right ∆ABC, ∠B = 90° and ∠A = θ

4 tan θ = 3 ⇒ tan θ

Let BC = 3k and AB = 4k

Then, AC^{2 }=
AB^{2} + BC^{2} = 16k^{2} + 9k^{2} = 25k^{2}

In right ∆ABC, ∠B = 90° and ∠A = θ

AC^{2 }= AB^{2}
+ BC^{2} = b^{2} + a^{2}

In right ∆ABC, ∠B = 90° and ∠A = θ

Let BC = k and AB = 2k

Then, AC^{2 }=
AB^{2} + BC^{2} = (4k^{2} + k^{2})= 5k^{2}

Let BC = 3k and AC = 4k

Then, AB^{2 }=
AC^{2} - BC^{2} = (16k^{2} - 9k^{2})= 7k^{2}

In right ∆ABC, ∠B = 90° and ∠A = θ

Let BC = 4k and AB = 3k

Then, AC^{2 }=
AB^{2} + BC^{2} = (9k^{2} + 16k^{2})= 25k^{2}

In right ∆ABC, ∠B = 90° and ∠A = θ

Let BC = 4k and AB = 3k

Then, AC^{2 }=
AB^{2} + BC^{2} = (9k^{2} + 16k^{2})= 25k^{2}

In right ∆ABC, ∠B = 90° and tan A = 1

Let BC = k and AB = k

Then, AC^{2 }=
AB^{2} + BC^{2} = (k^{2} + k^{2})= 2k^{2}

In right ∆ABC, ∠B = 90°

Then, BC^{2 }=
AC^{2} - AB^{2} = (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 = 120cm^{2}

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

Given:

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

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

By Pythagoras theorem, we have

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

Given:

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

By Pythagoras theorem, we have

Given:

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

By Pythagoras theorem, we have

Given:

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

By Pythagoras theorem, we have

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

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

## Chapter 10 - Trigonometric Ratios MCQ

Let BC = 8k and AB = 15k

Then, AC^{2} = AB^{2} + BC^{2} = (225k^{2} + 64k^{2}) = 289k^{2}

⇒ AC^{2} =289k^{2}

**Correct Option:** C

tan θ

Let BC = √3k and AB = k

Then, AC^{2}
= AB^{2} + BC^{2} = (k^{2}
+ 3k^{2}) = 4k^{2}

⇒
AC^{2} =4k^{2}

**Correct
Option:** A

cosec θ = √10

Let BC = k and AC = √10k

Then, AB^{2}
= AC^{2} - BC^{2} = (10k^{2}
- k^{2}) = 9k^{2}

⇒
AB^{2} =9k^{2}

⇒ AB = √9k^{2}
= 3k

**Correct
Option:** D

**Correct
Option:** B

Let AB = 4k and AC = 5k

Then, BC^{2} = AC^{2} - AB^{2} = (25k^{2} - 16k^{2}) = 9k^{2}

⇒ BC = √9k^{2} = 3k

**Correct Option:** A

Let AB = 4k and AC = 5k

Then, BC^{2}
= AC^{2} - AB^{2} = (25k^{2}
- 16k^{2}) = 9k^{2}

⇒ BC = √9k^{2}
= 3k

**Correct
Option:** A

**Correct Option:** D

We have

(tan θ + cot θ) = 5

⇒ (tan θ + cot θ)^{2}
= 5^{2}

⇒ tan^{2 }θ
+ cot^{2} θ + 2 tan θ cot θ = 25

⇒ tan^{2 }θ
+ cot^{2} θ + 2 tan θ ×

⇒ tan^{2 }θ
+ cot^{2} θ + 2 = 25

⇒ tan^{2 }θ
+ cot^{2} θ = 23

**Correct
Option:** D

We have,

(cos^{}θ +
sec θ) =

⇒ (cos^{}θ
+ sec θ)^{2} =

⇒ cos^{2 }θ
+ sec^{2} θ + 2 cos θ sec θ =

⇒ tan^{2 }θ
+ cot^{2} θ + 2 cos θ ×

⇒ tan^{2 }θ
+ cot^{2} θ + 2 =

⇒ tan^{2 }θ
+ cot^{2} θ =

**Correct
Option:** A

Let BC = 3k and AB = 4k

Then, AC^{2}
= AB^{2} + BC^{2} = (4k^{2}
+ 3k^{2}) = 25k^{2}

**Correct
Option:** B

**Correct
Option:** C

Let AB = 2k and AC = 3k

Then, BC^{2}
= AC^{2} - AB^{2} = (9k^{2}
- 4k^{2}) = 5k^{2}

**Correct
Option:** A

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

We know that,

(sec^{2} θ
- tan^{2} 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

cos A + cos^{2}
A = 1….given

⇒ cos A = (1 - cos^{2}
A) = sin^{2} A

∴ (sin^{2}
A + sin^{4} A) = (cos A + cos^{2} A) = 1

**Correct
Option:** C

**Correct Option:** D

**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

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