R S Aggarwal Class 10 Solutions for Chapter 10 Trigonometry Ratios are available here. Trigonometry Ratios is composed of only one exercise which includes concepts such as trigonometric ratios of an acute angle of a right triangle, reciprocal relations, T-ratios of an angle, quotient relation of T-ratios and squared relation. All textbook questions are solved by subject experts at BYJU’S based on CBSE Syllabus. Students can avail the R S Aggarwal Solutions for Chapter 10 now and clear their doubts on the chapter.

### Access Solutions to Maths R S Aggarwal Chapter 10 – Trigonometry Ratios

**Exercise 10 Page No: 539**

**Question 1:**

**Solution:**

Given function: sin θ = √3/2

Let us first draw a right ∆ABC, ∠B = 90 degrees and ∠A = 𝜃

(where k is a positive)

We know that sin 𝜃 = BC/AC = (Perpendicular)/Hypotenuse = √3/2

By Pythagoras Theorem:

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

Or AB^{2} = AC^{2} – BC^{2} = 4k^{2} – 3k^{2} = k^{2}

AB = k

**Find other T-rations using their definitions:**

Cos = AB/AC = 1/2

Tan 𝜃 = BC/𝐴𝐵 = √3

𝑐𝑜𝑠𝑒𝑐 𝜃 = 1/sin𝜃 = 2/√3

sec 𝜃 = 1/cos 𝜃 = 2

cot 𝜃 = 1/tan 𝜃 = 1/√3

**Question 2:**

**Solution:**

Given function: cos θ = 7/25

Draw a right ∆ABC, ∠B = 90 degrees and ∠A = θ

(where k is a positive)

We know that cos θ = AB/AC = Base/Hypotenuse = 7/25

By Pythagoras Theorem:

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

Or BC^{2} = AC^{2} – AB^{2} = 625k^{2} – 49k^{2} = 576k^{2}

AB = 24k

Find other T-rations using their definitions:

Sin 𝜃 = BC/AC = 24/25

Tan 𝜃 = BC/𝐴𝐵 = 24/7

𝑐𝑜𝑠𝑒𝑐 𝜃 = 1/sin𝜃 = 25/24

sec 𝜃 = 1/cos 𝜃 = 25/7

cot 𝜃 = 1/tan 𝜃 = 7/24

**Question 3:**

**Solution:**

Given function: tan θ=15/8

Draw a right ∆ABC, ∠B = 90 degrees and ∠A = θ

We know that tan θ = BC/AB = perpendicular/base = 15/8

(where k is a positive)

By Pythagoras Theorem:

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

= 64k^{2} + 225k^{2}

= 289k^{2}

AC = 17k

Find other T-rations using their definitions:

Sin 𝜃 = BC/AC = 15/17

cos 𝜃 = AB/AC = 8/17

𝑐𝑜𝑠𝑒𝑐 𝜃 = 1/sin𝜃 = 17/15

sec 𝜃 = 1/cos 𝜃 = 17/8

cot 𝜃 = 1/tan 𝜃 = 8/15

**Question 4:**

**Solution:**

Given function: cot θ= 2

Draw a right ∆ABC, ∠C = 90 degrees and ∠A = θ

We know that cot θ = AC/BC = base/perpendicular = 2/1

(where k is a positive)

By Pythagoras Theorem:

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

= k^{2} + 4k^{2}

= 5k^{2}

AB = k√5

Find other T-rations using their definitions:

Sin 𝜃 = BC/AB = k/(k√5) = 1/√5

cos 𝜃 = AC/AB = (2k)/(k√5) = 2/√5

tan θ = BC/AC = sinθ /cosθ = k/(2k) = 1/2

𝑐𝑜𝑠𝑒𝑐 𝜃 = 1/sin𝜃 = √5

sec 𝜃 = 1/cos 𝜃 = √5/2

**Question 5:**

**Solution:**

Given function: cosec θ = √10

Draw a right ∆ABC, ∠C = 90 degrees and ∠A = θ

We know that, cosecθ = AB/BC = hypotenuse/perpendicular = (k√10)/k

(where k is a positive)

By Pythagoras Theorem:

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

AB^{2} = AC^{2} – BC^{2}

= 10k^{2} – k^{2}

= 9k^{2}

AC = 3k

Find other T-rations using their definitions:

Sin 𝜃 = BC/AB = 1/√10

cos 𝜃 = AC/AB = (3k)/(k√10) = 3/√10

tan θ = BC/AC = sinθ /cosθ = 1/3

secθ = AB/AC = 1/cosθ = √10/3

cotθ = AC/BC = 1/tanθ = 3

**Question 6:**

**Solution:**

Draw a triangle, ΔABC, Let ∠ACB = θ and ∠B = 90 degrees

Sin θ = (a^{2} – b^{2}) / (a^{2} + b^{2})

AB = (a^{2} – b^{2})

AC = (a^{2} + b^{2})

By Pythagoras theorem:

BC = √[(a^{2} + b^{2})^{2} – (a^{2} – b^{2})^{2}]

BC = √(4a^{2} b^{2})

or BC = 2ab

Find other T-rations using their definitions:

cos 𝜃 = base/hypotenuse = 2ab / (a^{2} + b^{2})

tan θ = perpendicular/base = (a^{2} – b^{2}) / 2ab

cosec 𝜃 = 1/sin 𝜃 = (a^{2} + b^{2})/(a^{2} – b^{2})

sec θ = 1/cos 𝜃 = (a^{2} + b^{2})/2ab

cotθ = 1/tanθ = 2ab/(a^{2} – b^{2})

**Question 7:**

**Solution:**

Given: 15 cot A = 8

cotA = (8k)/(15k) = 1/tanA = AC/BC

Where k is any positive.

By Pythagoras theorem:

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

= (15k)^{2} + (8k)^{2}

= 289k^{2}

AB = 17k

Find other T-rations using their definitions:

sin A = perpendicular/ hypotenuse = (15k)/(17k) = 15/17

sec A = hypotenuse /base = 17/8

**Question 8:**

**Solution:**

Draw a triangle, ΔABC, Let ∠ACB = θ and ∠B = 90 degress

Sin A = perpendicular/ hypotenuse = 9/41

By Pythagoras theorem:

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

AB^{2} = AC^{2} – BC^{2}

= 41^{2} – 9^{2}

= 1600

AB = 40

Find other T-ratios using their definitions:

cos A = base/ hypotenuse = 40/41

tan A = perpendicular /base = 9/40

**Question 9:**

**Solution:**

cos θ = 0.6 = (6k)/(10k) = AC/AB

Where k is any positive.

By Pythagoras theorem:

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

BC^{2} = AB^{2} – AC^{2}

= (10k)^{2} + (6k)^{2}

= 64k^{2}

BC = 8k

Find other T-rations using their definitions:

sin θ = perpendicular/ hypotenuse = 8/10

tan θ = perpendicular/base = 8/6

Now,

LHS = 5sinθ – 3tanθ

= 5(8/10) – 3(8/6)

= 4 – 3(4/3)

= 4(3) – 3(4)

= 12 – 12

= 0

=RHS

Hence proved.

**Question 10:**

**Solution:**

cosec θ = 2

or 1/sinθ = 2

(cosecθ is reciprocal of sin θ)

sin θ = 1/2

which implies θ = 30 degrees.

Find the values of cos θ and cot θ at θ = 30 degrees.

cos 30^0 = √3/2 and cot 30^0 = √3

Now,

LHS = cot 0 + sin0/(1 +cos0)

= √3 + 1/2 / (1 + √3/2)

=2

=RHS

Hence proved.

**Question 11:**

**Solution:**

Given: tan θ = 1/√7

tanθ = k/(k√7) = BC/AC

Where k is any positive.

By Pythagoras theorem:

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

= k^{2} + 7k^{2}

AB = 2k√2

Find cosec θ and sec θ using their definitions:

cosec θ = AB/BC = 2k√2/k = 2√2

secθ = AB/AC = 2√2/√7

Now,

LHS =

= 48/64

=3/4

= RHS

Hence proved

**Question 12.**

**Solution:**

Given: tan θ = 20/21

tanθ = 20k/(21k)

Where k is any positive.

By Pythagoras theorem:

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

= 441k^{2} + 400k^{2}

AC = 29k

Find sin θ and cos θ using their definitions:

sin θ = perpendicular/ hypotenuse = 20/29

cos 𝜃 = base/hypotenuse = 21/29

Now,

LHS =

=30/70

= 3/7

=RHS

Hence proved

**Question 13:**

**Solution:**

Given: secθ=5/4

sec θ = 5k/(4k) = 5/4 and cos θ = 4/5

Where k is any positive.

By Pythagoras theorem:

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

BC^{2} = AC^{2} – AB^{2}

= 25k^{2} – 16k^{2}

BC = 9k^{2} = 3k

Find sin θ, tan θ and cot θ using their definitions:

sin θ = perpendicular/ hypotenuse = 3/5

tan θ = perpendicular/base = 3/4

cotθ = 1/tanθ = 4/3

Now,

LHS =

=RHS

**Question 14:**

**Solution:**

Given: cotθ=3/4

or cotθ = 3k/4k

Where k is any positive.

By Pythagoras theorem:

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

= 9k^{2} +16k^{2}

= 25k^{2}

AC = 5k

Find sec θ and cosec θ using their definitions:

sec θ = hypotenuse/base = 5/3

cosec 𝜃 = hypotenuse/perpendicular = 5/4

Now,

**Question 15:**

**Solution:**

Given: sinθ=3/4

or sinθ=3k/4k

Where k is any positive.

By Pythagoras theorem:

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

16k^{2} = AB^{2} + 9k^{2}

AB = √7

Find sec θ and cot θ using their definitions:

sec θ = hypotenuse/base = 4/√7

cotθ = base/perpendicular = √7/3

Now,

**Question 16:**

**Solution:**

Given: sinθ = a/b

or sinθ= ak/bk

Where k is any positive.

By Pythagoras theorem:

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

AB^{2} = AC^{2} – BC^{2}

= b^{2} – a^{2}

Find sec θ and tan θ using their definitions:

sec θ = hypotenuse/base = b/√(b^{2} – a^{2})

tan θ = perpendicular/base = a/√(b^{2} – a^{2})

Now,

LHS = sec θ + tan θ

= RHS

**Question 17:**

**Solution:**

Given: cosθ=3/5

cosθ = (3k)/(5k) = AC/AB

Where k is any positive.

By Pythagoras theorem:

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

BC = 4k

Find sin θ, tan θ and cot θ using their definitions:

sin θ = perpendicular/ hypotenuse = 4/5

tan θ = perpendicular/base = 4/3

cot θ = 1/tan θ = 3/4

Now,

LHS =

= 3/160

= RHS

**Question 18:**

Solution:

Given: tan θ = 4/3

tanθ = (4k)/(3k) = BC/AC

Where k is any positive.

By Pythagoras theorem:

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

AB^{2} = 16k^{2} + 9k^{2}

AB = 5k

Find sin θ and cos θ using their definitions:

sin θ = perpendicular/ hypotenuse = 4/5

cos 𝜃 = base/hypotenuse = 3/5

Now,

LHS = sin θ + cos θ

= 4/5 + 3/5

= 7/5

= RHS

**Question 19:**

**Solution:**

Given: tan θ = a/b

tanθ = (ak)/(bk) = BC/AB

Where k is any positive.

By Pythagoras theorem:

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

Find sin θ and cos θ using their definitions:

sin θ = perpendicular/ hypotenuse = a/(√(a^{2}+b^{2})

cos 𝜃 = base/hypotenuse = b/(√(a^{2}+b^{2})

Now,

LHS:

**Question 20:**

**Solution:**

Given: 3 tan θ = 4

or tan θ = 4/3

**Question 21:**

**Solution:**

Given: 3 cot θ = 2

or cot θ = 2/3

Where k is any positive.

By Pythagoras theorem:

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

AC^{2} = (2k)^{2} + (3k)^{2}

AC^{2} = 4k^{2} + 9k^{2} = 13k^{2}

AC = √13 k

Find sin θ and cos θ using their definitions:

sin θ = perpendicular/ hypotenuse = 3/√13

cos 𝜃 = base/hypotenuse =2/√13

Now,

**Question 22:**

**Solution:**

Given: 3cotθ=4

or cot θ = 4/3

Where k is any positive.

By Pythagoras theorem:

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

= 16 + 9

AC = 5 k

Find sin θ and cos θ using their definitions:

sin θ = perpendicular/ hypotenuse = 3/5

cos 𝜃 = base/hypotenuse = 4/5

Now,

**Question 23:**

**Solution:**

Given: sec θ = 17/8

or sec θ = 17k/8k

Where k is any positive.

By Pythagoras theorem:

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

BC^{2} = AC^{2} – AB^{2}

= 289k^{2} – 64k^{2}

= 225 k^{2}

BC = 15 k

**Question 24:**

**Solution:**

Draw a triangle using given instructions:

From figure: Δ ABC and Δ ABD are right angled triangles

where AD = 10cm BC = CD = 4cm

BD = BC + CD = 8cm

By Pythagoras theorem:

AD^{2} = BD^{2} + AB^{2}

(10)^{2} = (8)^{2} + AB^{2}

100 = 64 + AB^{2}

AB^{2} = 36 = (6)2

or AB = 6cm

Again,

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

AC^{2} = (4)^{2} + (6)^{2}

AC^{2} = 16 + 36 = 52

or AC = √52 = 2√13 cm

(i) Find sin θ

sin θ = BC/AC = 4/2√13 = 2√13/13

(ii) Find cos θ

cosθ = AB/AC = 6/2√13 = 3/√13 = 3√13/13

**Question 25:**

**Solution:**

Draw a triangle using given instructions:

From figure: Δ ABC is a right angled triangle

By Pythagoras theorem:

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

AC = 25

(i) Find sin A

sin A = BC/AC = 7/25

(ii) Find cos A

cos A = AB/AC = 24/25

(iii) sin C = AB/AC = 24/25

(iv) cosC = BC/AC = 7/25

**Question 26:**

**Solution:**

Draw a triangle using given instructions:

From figure: Δ ABC is a right angled triangle

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

(29)^{2} = AC^{2} + (21)^{2}

841 = AC^{2} + 441

AC^{2} = 400

or AC = 20

Find sin θ and cos θ:

sinθ = AC/AB = 20/29

cosθ = BC/AB = 21/29

Now:

LHS = cos^{2}θ – sin^{2}θ

= (21/29)^{2} – (20/29)^{2}

= 41/841

= RHS

Hence proved.

**Question 27:**

**Solution:**

From figure: Δ ABC is a right angled triangle

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

AC^{2} = (5)^{2} + (12)^{2}

AC^{2} = 25 + 144

AC^{2} = 169 = (13)^{2}

or AC = 13

Now from figure,

i. cos A = AB/AC = 12/13

ii. cosec A = AC/BC = 13/5

iii. cos C = BC/AC = 5/13

iv. cosec C = AC/AB = 13/12

Hence proved.

**Question 28:**

**Solution:**

Given: sinα = k/(2k) = BC/AB

Where k is any positive.

By Pythagoras theorem:

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

AC^{2} = AB^{2} – BC^{2}

= (2k)^{2} – (k)^{2}

= 3k^{2}

or AC= k√3

Find cos α:

cos α = base/hypotenuse = √3/2

Now,

**Question 29:**

**Solution:**

Given: tan A = BC/AB = k/(k√3)

Where k is any positive.

By Pythagoras theorem:

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

= (k)^{2} – (√3 k)^{2}

= k^{2} + 3k^{2}

= 4k^{2}

or AC= 2k

Find sin A, cos A, sin C and cos C

sin A = BC/AC = k/(2k) = 1/2

cos A = AB/AC = (k√3)/(2k) = √3/2

sin C = AB/AC = (k√3)/(2k) = √3/2

cos C = BC/AC = k/(2k) = ½

(i) sinA cosC + cosA sinC = 1

LHS = sinA cosC + cosA sinC = (1/2)(1/2) + ( √3/2)( √3/2)

= 1/4 + 3/4

= 4/4

= 1

RHS = LHS

(ii) cosA cosC – sinA sinC = 0

LHS = cosA cosC – sinA sinC

= (1/2)(√3/2) – (1/2)(√3/2)

= (√3/4) – (√3/4)

= 0

=RHS

**Question 30:**

**Solution:**

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

To Prove: ∠A = ∠B

From figures:

sin A = XY/XA and sin B = WZ / WB

XY/XA = WZ / WB = k (say)

or XY/WZ = XA/ WB …(1)

sin A = sin B (Given)

We have,

XY = WZ k and XA = WB k …(2)

By Pythagoras: Apply on both the triangles

WB^{2} = WZ^{2} + BZ^{2}

BZ^{2} = WB^{2} – WZ^{2}

And,

XA^{2} = XY^{2} + AY^{2}

AY^{2} = XA^{2} – XY^{2}

**Question 31:**

**Solution:**

Consider ΔABC to be a right angled triangle.

angle C = 90 degree

tan A = BC/AC and

tan B = AC/BC

Given: tanA = tanB

So, BC/AC = AC/BC

BC^{2} = AC^{2}

BC = AC

Which implies, ∠ A = ∠ B (using triangle opposite and equal angles property)

**Question 32:**

**Solution:**

Consider ΔABC to be a right angled triangle at B.

angle C = 90 degree

Given: tan A = 1 …(1)

tan A = 1 = BC/AB

AB = BC

Again, tan A = sin A/cos A

sin A = cos A …using (1)

By Pythagoras theorem:

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

AC^{2} = 2BC^{2}

(AC/BC)^{2} = 2

Or AC/BC = √2

cosecA = √2

or sin A = 1/√2

and cosA = 1/√2

Now,

2 sinA cosA = 2(1/√2)( 1/√2)

= 2(1/2)

= 1

= RHS

**Question 33:**

**Solution:**

Δ PQR is a right angled triangle.

By Pythagoras theorem:

PR^{2} = RQ^{2} + PQ^{2}

(x + 2)^{2} = x^{2} + PQ^{2}

PQ^{2} = 4 + 4x

or PQ = 2√(x+1)

Now,

cot ϕ = QR/PQ = x/2(√(x+1)

tan θ = QR/PQ = x/2(√(x+1)

(i) √(x+1) cot ϕ = √(x+1) {x/2(√(x+1)} = x/2

(ii) √(x^3 + x^ 2) tan θ = √(x^3 + x^ 2) {x/2(√(x+1)} = x^{2}/2

(iii) cos θ = PQ/PR = 2(√(x+1) / (x+2)

**Question 34:**

(Using trig property: sin^{2}** + **cos^{2} A = 1)

= 1 – 1

= 0

**Question 35:**

## R S Aggarwal Solutions for Chapter 10 Trigonometry Ratios Topics

In this chapter students will study important concepts on Trigonometry Ratios as listed below:

- Trigonometry Ratios introduction
- Trigonometric ratios of an acute angle of a right triangle
- Reciprocal relations
- T-ratios of an angle
- Quotient relation of T-ratios
- Squared relation

### Key Features of RS Aggarwal Solutions for Class 10 Maths Chapter 10 Trigonometry Ratios

1. A simple and easily understandable approach is followed to solve all the questions.

2. These solutions will be useful for CBSE board exams, and other competitive exams.

3. The step-by-step problem-solving approach helps students to understand easily.

4. Easy for quick revision.

5. Detailed answers to all the questions to help students in their preparations.