# RS Aggarwal Solutions Class 10 Chapter 5 Trigonometric Ratios

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## RS Aggarwal Solutions Class 10 Chapter 5 Trigonometric Ratios

### Exercise 5

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:

AC2 = AB2 + BC2

Or AB2 = AC2 – BC2 = 4k2 – 3k2 = k2

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:

AC2 = AB2 + BC2

Or BC2 = AC2 – AB2 = 625k2 – 49k2 = 576k2

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:

AC2 = AB2 + BC2

= 64k2 + 225k2

= 289k2

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:

AB2 = BC2 + AC2

= k2 + 4k2

= 5k2

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:

AB2 = AC2 + BC2

AB2 = AC2 – BC2

= 10k2 – k2

= 9k2

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 = (a2 – b2) / (a2 + b2)

AB = (a2 – b2)

AC = (a2 + b2)

By Pythagoras theorem:

BC = √[(a2 + b2)2 – (a2 – b2)2]

BC = √(4a2 b2)

or BC = 2ab

Find other T-rations using their definitions:

cos 𝜃 = base/hypotenuse = 2ab / (a2 + b2)

tan θ = perpendicular/base = (a2 – b2) / 2ab

cosec 𝜃 = 1/sin 𝜃 = (a2 + b2)/(a2 – b2)

sec θ = 1/cos 𝜃 = (a2 + b2)/2ab

cotθ = 1/tanθ = 2ab/(a2 – b2)

Question 7:
Solution:

Given: 15 cot A = 8

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

Where k is any positive.

By Pythagoras theorem:

AB2 = BC2 + AC2

= (15k)2 + (8k)2

= 289k2

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:

AC2 = AB2 + BC2

AB2 = AC2 – BC2

= 412 – 92

= 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:

AB2 = BC2 + AC2

BC2 = AB2 – AC2

= (10k)2 + (6k)2

= 64k2

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:

AB2 = BC2 + AC2

= k2 + 7k2

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:

AC2 = AB2 + BC2

= 441k2 + 400k2

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:

AC2 = AB2 + BC2

BC2 = AC2 – AB2

= 25k2 – 16k2

BC = 9k2 = 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:

AC2 = AB2 + BC2

= 9k2 +16k2

= 25k2

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:

AC2 = AB2 + BC2

16k2 = AB2 + 9k2

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:

AC2 = AB2 + BC2

AB2 = AC2 – BC2

= b2 – a2b2−a2−−−−−−√

Find sec θ and tan θ using their definitions:

sec θ = hypotenuse/base = b/√(b2 – a2)

tan θ = perpendicular/base = a/√(b2 – a2)

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:

AB2 = BC2 + AC2

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:

AB2 = BC2 + AC2

AB2 = 16k2 + 9k2

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:

AC2 = AB2 + BC2AC=a2+b2−−−−−−√

Find sin θ and cos θ using their definitions:

sin θ = perpendicular/ hypotenuse = a/(√(a2+b2)

cos 𝜃 = base/hypotenuse = b/(√(a2+b2)

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:

AC2 = AB2 + BC2

AC2 = (2k)2 + (3k)2

AC2 = 4k2 + 9k2 = 13k2

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:

AC2 = AB2 + BC2

= 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:

AC2 = AB2 + BC2

BC2 = AC2 – AB2

= 289k2 – 64k2

= 225 k2

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:

(10)2 = (8)2 + AB2

100 = 64 + AB2

AB2 = 36 = (6)2

or AB = 6cm

Again,

AC2 = BC2 + AB2

AC2 = (4)2 + (6)2

AC2 = 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:

AC2 = BC2 + AB2

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

AB2 = AC2 + BC2

(29)2 = AC2 + (21)2

841 = AC2 + 441

AC2 = 400

or AC = 20

Find sin θ and cos θ:

sinθ = AC/AB = 20/29

cosθ = BC/AB = 21/29

Now:

LHS = cos2θ – sin2θ

= (21/29)2 – (20/29)2

= 41/841

= RHS

Hence proved.

Question 27:
Solution:

From figure: Δ ABC is a right angled triangle

AC2 = BC2 + AB2

AC2 = (5)2 + (12)2

AC2 = 25 + 144

AC2 = 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:

AB2 = BC2 + AC2

AC2 = AB2 – BC2

= (2k)2 – (k)2

= 3k2

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:

AC2 = BC2 + AB2

= (k)2 – (√3 k)2

= k2 + 3k2

= 4k2

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

WB2 = WZ2 + BZ2

BZ2 = WB2 – WZ2

And,

XA2 = XY2 + AY2

AY2 = XA2 – XY2

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

BC2 = AC2

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:

AC2 = BC2 + AB2

AC2 = 2BC2

(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:

PR2 = RQ2 + PQ2

(x + 2)2 = x2 + PQ2

PQ2 = 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)} = x2/2

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

Question 34:

(Using trig property: sin2 + cos2 A = 1)

= 1 – 1

= 0

Question 35:

Complete RS Aggarwal Solutions Class 10

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