R S Aggarwal Solutions for Class 10 Maths Chapter 10 Trigonometry Ratios

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.

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RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10
RS Aggarwal Sol class 10 Maths Chapter 10

 

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

Exercise 10 Page No: 539

Question 1:

If sin θ = √3/2, find the value of all T-ratios of θ.

Solution:

Given function: sin θ = √3/2

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

rs aggarwal class 10 chapter 10 img 1

(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:

If cos θ=7/25 , find the values of all T-ratios of θ.

Solution:

Given function: cos θ = 7/25

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

rs aggarwal class 10 chapter 10 img 2

(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:

If tan θ=15/8, find the values of all T-ratios of θ.

Solution:

Given function: tan θ=15/8

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

rs aggarwal class 10 chapter 10 img 3

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:

If cot θ = 2, find the value of all T-ratios of θ.

Solution:

Given function: cot θ= 2

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

rs aggarwal class 10 chapter 10 img 4

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:

If cosec θ = √10, the find the values of all T-ratios of θ.

Solution:

Given function: cosec θ = √10

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

rs aggarwal class 10 chapter 10 img 5

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

(where k is a positive)

By Pythagoras Theorem:

AC2 = AB2 + 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:

If sinθ = (a2-b2)/(a2+b2), find the values of all T-ratios of θ.

Solution:

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

rs aggarwal class 10 chapter 10 img 6

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:

If 15cotA=8, find the values of sinA and secA.

Solution:

Given: 15 cot A = 8

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

rs aggarwal class 10 chapter 10 img 7

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:

If sinA=9/41, find the values of cosA and tanA.

Solution:

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

Sin A = perpendicular/ hypotenuse = 9/41

rs aggarwal class 10 chapter 10 img 8

By Pythagoras theorem:

AC2 = AB2 + BC2

AB2 = AC2 – BC2

= 412 – 92

= 1600

AB = 40

Find other T-rations using their definitions:

cos A = base/ hypotenuse = 40/41

tan A = perpendicular /base = 9/40

Question 9:

If cos θ = 0.6, show that (5sinθ − 3tanθ) = 0.

Solution:

cos θ = 0.6 = (6k)/(10k) = AC/AB

rs aggarwal class 10 chapter 10 img 9

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:

If cosec θ = 2, show that

rs aggarwal class 10 chapter 10 img 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)

rs aggarwal class 10 chapter 10 img 11

=2

=RHS

Hence proved.

Question 11:

If tan θ = 1/√7, show that

rs aggarwal class 10 chapter 10 img 12

Solution:

Given: tan θ = 1/√7

tanθ = k/(k√7) = BC/AC

rs aggarwal class 10 chapter 10 img 13

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

secθ = AB/AC = 2√2/√7

Now,

LHS =

rs aggarwal class 10 chapter 10 img 14

= 48/64

=3/4

= RHS

Hence proved

Question 12.

If tan θ = 20/21, show that

rs aggarwal class 10 chapter 10 img 15

Solution:

Given: tan θ = 20/21

tanθ = 20k/(21k)

rs aggarwal class 10 chapter 10 img 16

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 =

rs aggarwal class 10 chapter 10 img 17

rs aggarwal class 10 chapter 10 img 18

=30/70

= 3/7

=RHS

Hence proved

Question 13:

If secθ=5/4, show that

rs aggarwal class 10 chapter 10 img 19

Solution:

Given: secθ=5/4

sec θ = 5k/(4k) = 5/4 and cos θ = 4/5

rs aggarwal class 10 chapter 10 img 20

Where k is any positive.

By Pythagoras theorem:

AC2 = AB2 + BC2

BC2 = AC2 – AB2

= 25k2 + 16k2

BC = 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 =

rs aggarwal class 10 chapter 10 img 21

=RHS

Question 14:

If cotθ=3/4, show that

rs aggarwal class 10 chapter 10 img 22

Solution:

Given: cotθ=3/4

or cotθ = 3k/4k

rs aggarwal class 10 chapter 10 img 23

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,

rs aggarwal class 10 chapter 10 img 24

Question 15:

If sinθ=3/4, show that

rs aggarwal class 10 chapter 10 img 25

Solution:

Given: sinθ=3/4

or sinθ=3k/4k

rs aggarwal class 10 chapter 10 img 26

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,

rs aggarwal class 10 chapter 10 img 27

Question 16:

If sinθ=a/b, show that

rs aggarwal class 10 chapter 10 img 28

Solution:

Given: sinθ = a/b

or sinθ= ak/bk

rs aggarwal class 10 chapter 10 img 29

Where k is any positive.

By Pythagoras theorem:

AC2 = AB2 + BC2

AB2 = AC2 – BC2

= b2 – a2

\(\sqrt{b^{2}-a^{2}}\)

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

rs aggarwal class 10 chapter 10 img 30

= RHS

Question 17:

If cosθ=3/5, show that

rs aggarwal class 10 chapter 10 img 31

Solution:

Given: cosθ=3/5

cosθ = (3k)/(5k) = AC/AB

rs aggarwal class 10 chapter 10 img 32

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 =

rs aggarwal class 10 chapter 10 img 33

= 3/160

= RHS

Question 18:

If tan θ = 4/3, show that (sin θ + cos θ) = 7/5.

Solution:

Given: tan θ = 4/3

tanθ = (4k)/(3k) = BC/AC

rs aggarwal class 10 chapter 10 img 34

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:

If tan θ = a/b, show that

rs aggarwal class 10 chapter 10 img 35

Solution:

Given: tan θ = a/b

tanθ = (ak)/(bk) = BC/AB

rs aggarwal class 10 chapter 10 img 36

Where k is any positive.

By Pythagoras theorem:

AC2 = AB2 + BC2

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

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:

rs aggarwal class 10 chapter 10 img 37

Question 20:

If 3 tan θ = 4, show that

rs aggarwal class 10 chapter 10 img 38

Solution:

Given: 3 tan θ = 4

or tan θ = 4/3

rs aggarwal class 10 chapter 10 img 39

Question 21:

If 3 cot θ = 2, show that

rs aggarwal class 10 chapter 10 img 40

Solution:

Given: 3 cot θ = 2

or cot θ = 2/3

Where k is any positive.

By Pythagoras theorem:

AC2 = AB2 + BC2

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,

rs aggarwal class 10 chapter 10 img 41

Question 22:

If 3cotθ=4, show that (1-tan2 θ)/(1+tan2 θ) = (cos2 θ – sin2 θ).

 

Solution:

Given: 3cotθ=4

or cot θ = 4/3

rs aggarwal class 10 chapter 10 img 42

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,

rs aggarwal class 10 chapter 10 img 43

Question 23:

If sec θ = 17/8, verify that

rs aggarwal class 10 chapter 10 img 44

Solution:

Given: sec θ = 17/8

or sec θ = 17k/8k

rs aggarwal class 10 chapter 10 img 45

Where k is any positive.

By Pythagoras theorem:

AC2 = AB2 + BC2

BC2 = AC2 – AB2

= 289k2 – 64k2

= 225 k2

BC = 15 k

rs aggarwal class 10 chapter 10 img 46

Question 24:

In the adjoining figure, ∠B=90°, ∠BAC=θ°, BC=CD=4 cm and AD=10 cm.

Find (i) sinθ and (ii) cosθ.

Solution:

Draw a triangle using given instructions:

rs aggarwal class 10 chapter 10 img 47

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

where AD = 10cm BC = CD = 4cm

BD = BC + CD = 8cm

By Pythagoras theorem:

AD2 = BD2 + AB2

(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:

In a ∆ABC, it is given that ∠B = 90°, AB = 24 cm and BC = 7 cm.

Find the value of

(i) sin A

(ii) cos A

(iii) sin C

(iv) cos C

Solution:

Draw a triangle using given instructions:

rs aggarwal class 10 chapter 10 img 48

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:

In a ∆ABC, in which ∠C = 90°, ∠ABC = θ°, BC = 21 units, AB = 29 units.

Show that (cos2θ – sin2 θ) = 41/841.

Solution:

Draw a triangle using given instructions:

rs aggarwal class 10 chapter 10 img 49

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:

In a ∆ABC, angle B = 90 degrees, AB = 12 cm and BC = 5 cm.

Find (i) cos A (ii) cosec A (iii) cos C (iv) coses C

Solution:

rs aggarwal class 10 chapter 10 img 50

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:

If sinα=1/2, prove that (3cosα – 4cos^3α)=0.

Solution:

Given: sinα = k/(2k) = BC/AB

rs aggarwal class 10 chapter 10 img 51

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,

rs aggarwal class 10 chapter 10 img 52

Question 29:

In a ∆ABC, ∠B = 90° and tan A =1/√3. Prove that

(i) sinA·cosC + cosA·sinC = 1

(ii) cosA·cosC – sinA·sinC = 0

Solution:

Given: tan A = BC/AB = k/(k√3)

rs aggarwal class 10 chapter 10 img 53

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:

If ∠A and ∠B are acute angles such that sin A = sin B, then prove that ∠A = ∠B.

Solution:

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

rs aggarwal class 10 chapter 10 img 54

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

rs aggarwal class 10 chapter 10 img 55

Question 31:

If ∠A and ∠B are acute angles such that tan A = tan B, the prove that ∠A=∠B.

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:

In a right ∆ABC , right-angled at B, if tan A = 1 , then verify that 2sinA·cosA=1.

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:

In the figure of ∆PQR, ∠P=θ° and ∠R=ϕ°.

Find

(i) √(x+1) cot ϕ

(ii) √(x^3 + x^ 2) tan θ

(iii) cos θ

Solution:

rs aggarwal class 10 chapter 10 img 56

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:

If x = cosecA+cosA and y=cosecA-cosA , then prove that

rs aggarwal class 10 chapter 10 img 57

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

= 1 – 1

= 0

Question 35:

If x = cot A + cos A and y = cot A – cos A, prove that

rs aggarwal class 10 chapter 10 img 58

 

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.