Category: 10th mathematics exercise questions with answers

Trigonometry Exercise 13.4 – Class 10

Trigonometry Exercise 13.4 – Solutions:

  1. Evaluate:
  2. tan65˚/cot25˚
  3. sin18˚/cos72˚

iii. cos48˚- sin42˚

  1. cosec31˚ – sec59˚
  2. cot34˚ – tan56˚
  3. sin36˚/cos54˚sin54˚/cos36˚

vii. sec70˚ sin20˚ – cos70˚cosec20˚

viii. cos213˚ – sin277˚

 

  1. Prove that:
  2. sin35˚ sin55˚ – cos35˚cos55˚ = 0
  3. tan10˚tan15˚tan75˚tan80˚ = 1

iii. cos38˚cos52˚ – sin38˚sin52˚ = 0

 

III. If sin5θ = cos4θ, where 4θ and 5θ are acute angles, find the value of θ

 

  1. If sec4A = cosec(A – 20˚), where 4A is an acute angle, find the value of A.

Trigonometry Exercise 13.4 – Questions:

I. Evaluate:

  1. tan65˚/cot25˚

Solution:

We know,

cot25˚ = tan(90˚ – 25˚) = tan65˚

Therefore,  tan65˚/cot25˚ = tan65˚/tan65˚ =  1

 

 

ii. sin18˚/cos72˚

Solution:

We know,

cos72˚ = sin(90˚ – 72˚) = sin18˚

Therefore,  sin18˚/cos72˚ = sin18˚/sin18˚ =  1

 

iii. cos48˚- sin42˚

Solution:

We know,

cos48˚ = sin(90˚ – 48˚) = sin48˚

Therefore,  cos48˚- sin42˚= cos48˚- cos48˚ =  0

 

iv.  cosec31˚ – sec59˚

Solution:

We know,

cosec31˚ = sec(90˚ – 31˚) = sec59˚

Therefore,  cosec31˚ – sec59˚= cosec31˚ – cosec31˚=  0

 

v. cot34˚ – tan56˚

Solution:

We know,

cot34˚ = tan(90˚ – 34˚) = tan56˚

Therefore,  cot34˚ – tan56˚= cot34˚ – cot34˚=  0

 

vi. sin36˚/cos54˚sin54˚/cos36˚

Solution:

We know,

cos54˚ = sin(90˚ – 54˚) = sin36˚

cos36˚ = sin(90˚ – 36˚) = sin54˚

Therefore,  sin36˚/cos54˚sin54˚/cos36˚ = sin36˚/sin36˚sin54˚/sin54˚=  1 – 1  = 0

 

vii. sec70˚ sin20˚ – cos70˚cosec20˚

Solution:

We know,

sec70˚ = cosec(90˚ – 70˚) = cosec20˚

sec20˚ = cos(90˚ – 20˚) = cos70˚

Therefore,  sec70˚ sin20˚ – cos70˚cosec20˚ = cosec20˚ cos70˚ – cos70˚cosec20˚=  0

 

viii. cos213˚ – sin277˚

Solution:

We know, cos213˚ – sin277˚

cos213˚ = 1 – sin213˚

sin213˚ = cos2(90˚ – 13˚) = sin2(77˚)

Therefore,  cos213˚ – sin277˚ = 1 – sin277˚- sin277˚=  1


II. Prove that:

  1. sin35˚ sin55˚ – cos35˚cos55˚ = 0

Solution:

sin35˚ sin55˚ – cos35˚cos55˚ = 0

cos35˚ = sin(90˚ – 35˚) = sin55˚

cos55˚ = sin(90˚ –  55˚) =  sin35˚

Therefore, sin35˚ sin55˚ – cos35˚cos55˚

= sin35˚ sin55˚ – sin35˚ sin55˚

= 0

 

ii. tan10˚tan15˚tan75˚tan80˚ = 1

Solution:

sin10˚/cos10 . sin15˚/cos15. sin75˚/cos75. sin80˚/cos80 = 1

cos 10˚ = sin(90˚ – 10˚) = sin 80˚

cos 15˚ = sin(90˚ – 15˚) = sin75˚

cos75˚ = sin(90˚ – 75˚)  = sin15˚

cos80˚ = sin(90˚ – 10˚) = sin 10˚

⇒  sin10˚/sin80˚ . sin15˚/sin75˚. sin75˚/sin15˚. sin80˚/sin10˚ = 1

⇒ tan10˚tan15˚tan75˚tan80˚ = 1

 

iii. cos38˚cos52˚ – sin38˚sin52˚ = 0

Solution:

cos38˚cos52˚ – sin38˚sin52˚ = 0

we know,

cos38 = sin(90 – 38) = sin 52

cos52 = sin(90 – 52) = sin38

⇒ cos38˚cos52˚ – sin38˚sin52˚ = sin52˚sin38˚ – sin38˚sin52˚ = 0

 


III. If sin5θ = cos4θ, where 4θ and 5θ are acute angles, find the value of θ

Solution:

Given, sin5θ = cos4θ

We know, cos4θ = sin(90˚ – 4θ)

Since, sin(90˚ – 4θ) = sin5θ

⇒ 90˚ – 4θ = 5θ

⇒ 90˚ = 9θ

⇒ θ = 90˚/9 = 10˚

 


IV. If sec4A = cosec(A – 20˚), where 4A is an acute angle, find the value of A.

Solution:

Given  sec4A = cosec(A – 20˚)

sec4A = cosec(90˚ – 4A)

Since cosec(90˚ – 4A) = cosec(A – 20˚)

90˚ – 4A = A – 20˚

90˚ + 20˚ = A + 4A

110˚ = 5A

A = 110˚/5 = 22˚

Trigonometry Exercise 13.3 – Class 10

Trigonometry Exercise 13.3 – Questions:

I. Show that

  1. (1 – sin2θ) sec2 θ = 1
  2. (1 + tan2 θ) cos2 θ = 1
  3. (1 + tan2 θ)(1 – sin θ)( 1 + sin θ) = 1
  4. sin θ/(1+cosθ) + 1+cosθ/sinθ = 2 cosec θ
  5. 1 + sinθ/1 – sinθ = (sec θ + tan θ)2
  6. cosA/1-tanA + sinA/1 – cotA = sinA + cosA
  7. (1 – tan2A)/(1+tan2A) = 1 – 2sin2A
  8. (sinθ + cosθ)2 = 1 + 2sinθcosθ
  9. sinA cosA tanA + cosA.sinA.cotA = 1
  10. (tanA – sinA)/(sin2A) = tanA/ (1 + cosA)
  11. tan2A – sin2A = tan2A sin2A
  12. cos2A – sin2A = 2cos2A – 1

Trigonometry Exercise 13.3 – Solutions:

I. Show that

  1. (1 – sin2θ) sec2 θ = 1

Solution:

(1 – sin2θ) sec2 θ = 1

LHS =  (1 – sin2θ) sec2 θ

[since 1 – sin2θ  = cos2θ  and  sec2θ = 1/cos2θ]

= cos2θ .(1/cos2θ)

= 1

=  RHS

Thereore, (1 – sin2θ) sec2 θ = 1

 

  1. (1 + tan2 θ) cos2 θ = 1

Solution:

(1 + tan2 θ) cos2 θ = 1

LHS = (1 + tan2 θ) cos2 θ [since(1 + tan2 θ)  = sec2θ]

= sec2θ.cos2 θ

= (1/cos2θ).cos2 θ

=  1

= RHS

Therefore, (1 + tan2 θ) cos2 θ = 1

 

  1. (1 + tan2 θ)(1 – sin θ)( 1 + sin θ) = 1

Solution:

(1 + tan2 θ)(1 – sin θ)( 1 + sin θ) = 1

LHS = (1 + tan2 θ)(1 – sin θ)( 1 + sin θ)

= (1 + tan2 θ)(1 – sin2 θ) [since (a+b)(a-b) = a2 – b2]

= sec2θ.cos2θ

= (1/cos2θ).cos2θ

= 1

= RHS

Therefore, (1 + tan2 θ)(1 – sin θ)( 1 + sin θ) = 1

 

  1. sin θ/(1+cosθ) + 1+cosθ/sinθ = 2 cosec θ

Solution:

sin θ/(1+cosθ) + 1+cosθ/sinθ = 2 cosec θ

LHS = sin θ/(1+cosθ) + 1+cosθ/sinθ

= [sinθ.sinθ +(1+cos θ)(1+cos θ)]/[sin θ.(1+cos θ)]

= [sin2 θ+(1+cos θ)2]/[sin θ.(1+cos θ)]

=[sin2 θ + 1+ cos2 θ+2cos θ]/[sin θ(1+cos θ)]

=[2+2cos θ]/[sin θ(1+cos θ)]

=[2(1+cos θ)]/[sin θ(1+cos θ)]

=2/sin θ

= 2 cosec θ

= RHS

Therefore, sin θ/(1+cosθ) + 1+cosθ/sinθ = 2 cosec θ

 

  1. 1 + sinθ/1 – sinθ = (sec θ + tan θ)2

Solution:

1 + sinθ/1 – sinθ = (sec θ + tan θ)2

LHS = 1 + sinθ/1 – sinθ

= 1 + sinθ/1 – sinθ x 1 + sinθ/1 + sinθ

= (1 + sinθ)2/(1- sin2 θ)

= (1 + sinθ)2/(cos2 θ)

= [(1+sinθ)/cosθ]2

= [1/cosθ + sinθ/cosθ]2

= (sec θ + tan θ)2

= RHS

Therefore, 1 + sinθ/1 – sinθ = (sec θ + tan θ)2

 

  1. cosA/1-tanA + sinA/1 – cotA = sinA + cosA

Solution:

cosA/1-tanA + sinA/1 – cotA = sinA + cosA

LHS = cosA/1-tanA + sinA/1 – cotA

= cosA/1-(sinA/cosA) + sinA/1 – (cosA/sinA)

= cosA.cosA/cosA-sinA + sinA.sinA/sinA – cosA

= [cos2A/(cosA- sinA)] + [sin2A/(sinA – cosA)]

= [cos2A/(cosA- sinA)] – [sin2A/(cosA – sinA)]

= [cos2A – sin2A]/[cosA – sinA]

= [cosA+sinA][cosA-sinA]/[cosA-sinA]

= cosA + sinA

= RHS

Therefore, cosA/1-tanA + sinA/1 – cotA = sinA + cosA

 

  1. (1 – tan2A)/(1+tan2A) = 1 – 2sin2A

Solution:

(1 – tan2A)/(1+tan2A) = 1 – 2sin2A

LHS = (1 – tan2A)/(1+tan2A)

= [1 – (sin2A/cos2A)]/ [1 + (sin2A/cos2A)]

= [(cos2A – sin2A)/cos2A]/[(cos2A + sin2A)/cos2A]

=(cos2A – sin2A)/ (cos2A + sin2A)

= (1 –  sin2A – sin2A)/1

= 1 – 2sin2A

= RHS

Therefore, (1 – tan2A)/(1+tan2A) = 1 – 2sin2A

 

  1. (sinθ + cosθ)2 = 1 + 2sinθcosθ

Solution:

(sinθ + cosθ)2 = 1 + 2sinθcosθ

LHS = (sinθ + cosθ)2

= sin2 θ + cos2 θ + 2sinθ.cosθ

= 1 + 2sinθcosθ

= RHS

Therefore, (sinθ + cosθ)2 = 1 + 2sinθcosθ

 

  1. sinA cosA tanA + cosA.sinA.cotA = 1

Solution:

sinA cosA tanA + cosA.sinA.cotA = 1

LHS = sinA cosA tanA + cosA.sinA.cotA

= sinA cosA sinA/cosA + cosA.sinA. cosA/sinA

= sin2A + cos2A

= 1

= RHS

Therefore, sinA cosA tanA + cosA.sinA.cotA = 1

 

  1. (tanA – sinA)/(sin2A) = tanA/ (1 + cosA)

Solution:

(tanA – sinA)/(sin2A) = tanA/ (1 + cosA)

LHS = (tanA – sinA)/(sin2A)

= (sinA/cosA – sinA)/(sin2A)

= [sinA-sinA.cosA]/cosAsin2A

= sinA(1 – cosA)/cosA.(1 – cos2A) [since (a+b)(a-b) = a2 – b2]

= sinA(1 – cosA)/cosA.(1 – cosA)(1 + cosA)

= sinA/cosA (1 + cosA)

= tanA/(1+cosA)

= RHS

Therefore, (tanA – sinA)/(sin2A) = tanA/ (1 + cosA)

 

  1. tan2A – sin2A = tan2A sin2A

Solution:

tan2A – sin2A = tan2A sin2A

LHS = tan2A – sin2A

=[sin2A/cos2A] – sin2A

= [(sin2A – sin2A.cos2A)/cos2A] – sin2A

= [sin2A(1 – cos2A)]/cos2A

=sin2A.sin2A/cos2A

= tan2A sin2A

= RHS

Therefore, tan2A – sin2A = tan2A sin2A

 

  1. cos2A – sin2A = 2cos2A – 1

Solution:

cos2A – sin2A = 2cos2A – 1

LHS = cos2A – sin2A

= cos2A – (1 – cos2A)

= cos2A – 1 + cos2A)

= 2cos2A – 1

= RHS

Therefore, cos2A – sin2A = 2cos2A – 1

Trigonometry Exercise 13.2 – Class X

Trigonometry Exercise 13.2 – Questions:

I. Answer the following questions:

  1. What trigonometric ratios of angles from 0 to 90 are equal to 0?
  2. Which trigonometric ratios of angles from 0 to 90 are equal to 1?
  3. Which trigonometric ratios of angles from 0 to 90 are equal to 1/2?
  4. Which trigonometric ratios of angles from 0 to 90 are not defined?
  5. Which trigonometric ratios of angles from 0 to 90 are equal?

II. Find θ. if 0≤θ≤90

  1. √2 cos θ = 1
  2. √3 tan θ = 1
  3. 2 sin θ = √3
  4. 5 sin θ = 0
  5. 3 tan θ = √3

III. Find the value of the following.

  1. sin 30˚ cos 60˚ – tan245˚
  2. sin 60˚ cos 30˚ + cos 60˚ sin 30˚
  3. cos 60˚ cos 30˚ – sin 60˚ sin 30˚
  4. 2sin230˚ – 3 cos2 30˚ + tan 60˚ + 3 sin2 90˚
  5. 4 sin2 60˚ + 3 tan2 30˚ – 8 sin 45˚ .cos 45˚
  6. cos 45˚/sec 30˚ + cosec 30˚
  7. (4sin260 – cos245)/(tan230 + sin20)
  8. (sin30˚ + tan45˚ – cosec60˚)/(sec30˚ + cos 60˚ +cot45˚)
  9. (5cos260˚ + 4 sec230˚ – tan245˚)/(sin230˚ + cos230˚)
  10. (5sin230˚ + cos245˚ – 4 tan230˚)/(2sin30˚ + cos30˚ + tan45˚)

IV. Prove the following equalities.

  1. sin30˚ .cos60 ˚+ cos30˚.sin60˚ = sin 90˚
  2. 2cos230˚ – 1 = 1 – 2sin230˚ = cos 60˚
  3. if θ = 30˚, prove that 4cos2θ – 3 cos θ = cos 3θ
  4. If π = 180˚ and A = π/6 prove that (1 – cosA)(1 + cos A)/(1 – sin A)(1+ sinA) = 1/3
  5. If B = 15˚, prove that 4 sin 2B.cos 4B.sin 6B = 1
  6. If A = 60˚ and B = 30˚ then prove that tan(A – B) = tanA – tanB/1 + tanAtanB

Trigonometry Exercise 13.2 – Solutions:

I. Answer the following questions:

  1. What trigonometric ratios of angles from 0 to 90 are equal to 0?

Solution:

sinθ = 0 when θ = 0˚

cosθ = 0 when θ = 90˚

tanθ = 0 when θ = 0˚

cotθ = 0 when θ = 90˚

 

  1. Which trigonometric ratios of angles from 0 to 90 are equal to 1?

Solution:

sinθ = 1 when θ = 90˚

cosθ = 1 when θ = 0˚

tanθ = 1 when θ = 45˚

cosec θ  = 1  when  θ = 90 ˚

sec θ = 1 when θ = 0 ˚

 

  1. Which trigonometric ratios of angles from 0 to 90 are equal to 1/2?

Solution:

sinθ = 1/2 when θ = 30˚

cosθ = 1/2 when θ = 60˚

 

  1. Which trigonometric ratios of angles from 0 to 90 are not defined?

Solution:

tanθ = undefined when θ = 90˚

cosecθ = undefined when θ = 0˚

secθ = undefined when θ = 90˚

cotθ = undefined when θ = 0˚


  1. Find θ. if 0≤θ≤90
  2. √2 cos θ = 1

Solution:

√2 cos θ = 1

cos θ = 1/√2

We know, cos 45˚ = 1/√2

Therefore, cos θ = cos 45˚

Thus, θ = 45˚

 

  1. √3 tan θ = 1

Solution:

√3 tan θ = 1

tan θ = 1/√3

We know, tan 30˚ = 1/√3

Therefore, cos 30˚ = cos θ

Thus, θ = 30˚

 

  1. 2 sin θ = √3

Solution:

2 sin θ = √3

sin θ = √3/2

We know, sin 60˚ = √3/2

Therefore, sin 60˚ = sin θ

Thus, θ = 60˚

 

  1. 5 sin θ = 0

Solution:

5 sin θ = 0

sin θ = 0/5

sin θ = 0

We know, sin 0˚ = 0

Therefore, sin 0˚ = sin θ

Thus, θ = 0˚

 

  1. 3 tan θ = √3

Solution:

3 tan θ = √3

tan θ = √3/3

i.e., tan θ = 1/√3

We know, tan 30˚ = 1/√3

Therefore, tan 30˚ = tan θ

Thus, θ = 30˚


III. Find the value of the following.

  1. sin 30˚ cos 60˚ – tan245˚

Solution:

sin 30˚ cos 60˚ – tan245˚

= 1/2 x 1/2 – (1)2

= 1/4 – 1

= -3/4

 

  1. sin 60˚ cos 30˚ + cos 60˚ sin 30˚

Solution:

sin 60˚ cos 30˚ + cos 60˚ sin 30˚

= √3/2 x √3/2 + 1/2 x 1/2

= 3/4 + 1/4

= 3+1/4

= 4/4

= 1

 

  1. cos 60˚ cos 30˚ – sin 60˚ sin 30˚

Solution:

cos 60˚ cos 30˚ – sin 60˚ sin 30˚

= 1/2 x √3/2  – √3/2 x 1/2

= √3/4√3/2

= 0

 

  1. 2sin230˚ – 3 cos2 30˚ + tan 60˚ + 3 sin2 90˚

Solution:

2sin230˚ – 3 cos2 30˚ + tan 60˚ + 3 sin2 90˚

= 2(1/2)2 – 3(√3/2 )2 + (√3) + 3(1)2

= 2(1/4) – 3(3/4) + √3 + 3

= 1/29/4 + √3 + 3

= 5/4 + √3

 

  1. 4 sin2 60˚ + 3 tan2 30˚ – 8 sin 45˚ .cos 45˚

Solution:

4 sin2 60˚ + 3 tan2 30˚ – 8 sin 45˚ .cos 45˚

= 4(√3/2)2 + 3(1/√3)2 – 8(1/√2) .( 1/√2)

= 4(3/4) + 3(1/3) – 8(1/2)

= 3 + 1 – 4

= 0

 

  1. cos 45˚/(sec 30˚ + cosec 30˚)

Solution:

cos 45˚/(sec 30˚ + cosec 30˚)

= (1/√2)/( 2/√3+ 2)

= (1/√2)/(2+2√3/√3)

√3/2√2(1+√3)

 

7.

Trigonometry Exercise 13.2

 

  1. (sin30˚ + tan45˚ – cosec60˚)/(sec30˚ + cos 60˚ +cot45˚)

Solution:

(sin30˚ + tan45˚ – cosec60˚)/(sec30˚ + cos 60˚ +cot45˚)

=(1/2  +1 –  2/√3)/(2/√3 + 1/2 + 1)

= (9  – 4√3/6)/(9 + 4√3/6)

9  – 4√3/9  – 4√3

 

  1. (5cos260˚ + 4 sec230˚ – tan245˚)/(sin230˚ + cos230˚)

Solution:

(5cos260˚ + 4 sec230˚ – tan245˚)/(sin230˚ + cos230˚)

= [5(1/2)2 + 4(2/√3)2 – (1)2]/[(1/2)2 + (√3/2)2]

= [5/4 + 16/3 – 1]/[1/4 + 3/4]

= (67/12)/(1)

= 67/12

 

  1. (5sin230˚ + cos245˚ – 4 tan230˚)/(2sin30˚ + cos30˚ + tan45˚)

Solution:

(5sin230˚ + cos245˚ – 4 tan230˚)/(2sin30˚ + cos30˚ + tan45˚)

= [5(1/2)2 + (1/√2)2 – 4(1/√3)2]/[2(1/2) + (√3/2) + 1]

= [5/4 + 1/24/3]/[1 + √3/2 + 1]

= [5/12]/[4+√3/2]

= 5/6(4+√3)

 


IV.Prove the following equalities.

  1. sin30˚ .cos60 ˚+ cos30˚.sin60˚ = sin 90˚

Solution:

sin30˚ .cos60 ˚+ cos30˚.sin60˚ = sin 90˚

LHS = sin30˚ .cos60 ˚+ cos30˚.sin60˚

= 1/2 . 1/2 + √3/2.√3/2

= 1/4 + 3/4

= 1+3/4

= 1

RHS = sin 90˚ = 1

Therefore, LHS = RHS

 

  1. 2cos230˚ – 1 = 1 – 2sin230˚ = cos 60˚

Solution:

2cos230˚ – 1 = 1 – 2sin230˚ = cos 60˚

Take, 2cos230˚ – 1 ………………..(1)

= 2(√3/2)2 – 1

= 2(3/4) – 1

= 3/2 – 1

= 3 – 2/2 = 1/2

1 – 2 sin2 30˚ ………………(2)

= 1 – 2(1/2)2

= 1 – 2(1/4)

= 1 – 1/2

= 1/2

cos60˚ …………………(3)

we know, cos60˚ = 1/2

Therefore, (1) = (2) = (3)

Thus, 2cos230˚ – 1 = 1 – 2sin230˚ = cos 60˚

 

  1. If θ = 30˚, prove that 4cos3θ – 3 cos θ = cos 3θ

Solution:

If θ = 30˚, we have to prove that 4cos3θ – 3 cos θ = cos 3θ

LHS = 4cos3θ – 3 cos θ

= 4cos330˚ – 3 cos 30˚

= 4(√3/2)3 – 3(√3/2)

= 4(3√3/8) – 3(√3/2)

= 3√3/23√3/2

= 3√3 – 3√3/2

= 0

RHS = cos 3θ

= cos 3(30˚)

= cos(90˚)

= 0

LHS = RHS

 

  1. If π = 180˚ and A = π/6 prove that (1 – cosA)(1 + cos A)/(1 – sin A)(1+ sinA) = 1/3

Solution:

If π = 180˚ and A = π/6 = 30˚ we have to prove that (1 – cosA)(1 + cos A)/(1 – sin A)(1+ sinA) = 1/3

LHS = (1 – cosA)(1 + cos A)/(1 – sin A)(1+ sinA)

= (1 – cos 30˚)(1 + cos 30˚)/(1 – sin 30˚)(1+ sin30˚)

= [1 – cos230˚]/[1 – sin230˚]

= [1 – (√3/2)2]/[1 – (1/2)2]

= [1 – 3/4]/[1 – 1/4]

= [4 – 3/4]/[4-1/4]

=[ 1/4]/[3/4]

= 1/3

= RHS

 

  1. If B = 15˚, prove that 4 sin 2B.cos 4B.sin 6B = 1

Solution:

If B = 15˚, we have to prove that 4sin 2B.cos 4B.sin 6B = 1

LHS = 4sin 2B.cos 4B.sin 6B

= 4sin 2(15˚).cos 4(15˚).sin 6(15˚)

= 4 sin(30˚).cos(60˚).sin(90˚)

= 4(1/2).(1/2).(1)

= 4(1/4)

= 1

= RHS

 

  1. If A = 60˚ and B = 30˚ then prove that tan(A – B) = tanA – tanB/1 + tanAtanB

Solution:

If A = 60˚ and B = 30˚ then we have to prove that tan(A – B) = tanA – tanB/1 + tanAtanB

LHS = tan(A – B)

= tan(60˚ – 30˚)

= tan30˚

= 1/√3

RHS = tanA – tanB/1 + tanAtanB

= tan60˚ – tan30˚/1 + tan60˚tan30˚

= [√3 – 1/√3]/[1 + √3.(1/√3)]

= (2√3/3)/(1+1)

= √3/3

= 1/√3

Therefore, LHS = RHS

Trigonometry Exercise 13.1 – Class 10

Trigonometry Exercise 13.1 – Questions:

  1. Find sin θ and cos θ for the following:

(i)

Trigonometry Exercise 13.1

(ii)

Trigonometry Exercise 13.1

(iii)

Trigonometry Exercise 13.1

2.Find the following:

  1. If sin x = 3/5 , cosec x = ______________
  2. If cos x = 24/25, sec x = _______________
  3. If tan x = 7/24 , cot x = _______________
  4. If cosec x = 25/15 , sin x = ____________
  5. If sin A = 3/5 and cos A = 4/5, then, tan A = __________
  6. If cot A = 8/15 and sin A = 15/17, then, cos A = _________

 

3.Solve:

  1. Given tan A = 3/4, find the value of sin A and cos A.
  2. Given cot θ = 20/21­, determine cos θ and cosec θ
  3. Given tan A = 7/24­, find the other trigonometric ratios of angle A.
  4. If 2 sin θ = √3, find cos θ, tan θ and cot θ + cosec θ
  5. If 3 tan θ = 1, find sin θ, cos θ and cot θ
  6. If sec x = 2, then, find sin x, tan x, cot x and cot x + cosec x
  7. If 4 sin A – 3 cos A = 0, find sin A, cos A, sec A and cosec A
  8. If 13 sin A = 5 and A is acute, find the va;ue of 5sin A – 2 cos A/tan A
  9. If cos θ = 5/13 and θ is acute, find the value of 5tanθ + 12cot θ/5tanθ – 12cot θ
  10. IF 13 cos θ – 5 = 0, find sin θ + cos θ/sin θ – cos θ

Trigonometry Exercise 13.1 – Solutions:

  1. Find sinθ and cosθ for the following:

(i)

Trigonometry Exercise 13.1

(ii)

Trigonometry Exercise 13.1

(iii)

Trigonometry Exercise 13.1

Solution:

Trigonometry Exercise 13.1(i)

sin θ = opposite side/hypotensuse = 24/25

cos θ = adjacent side/hypotensuse = 7/25

 

 

Trigonometry Exercise 13.1(ii)

sin θ = opposite side/hypotensuse = 15/25

cos θ = adjacent side/hypotensuse = 20/25

 

 

Trigonometry Exercise 13.1(iii)

sin θ = opposite side/hypotensuse = 10/25

cos θ = adjacent side/hypotensuse = 24/25

 

 


2.Find the following:

  1. If sin x = 3/5 , cosec x = ______________
  2. If cos x = 24/25, sec x = _______________
  3. If tan x = 7/24 , cot x = _______________
  4. If cosec x = 25/15 , sin x = ____________
  5. If sin A = 3/5 and cos A = 4/5, then, tan A = __________
  6. If cot A = 8/15 and sin A = 15/17, then, cos A = _________

Solution:

  1. If sin x = 3/5 , cosec x = 5/3
  2. If cos x = 24/25, sec x = 25/24
  3. If tan x = 7/24 , cot x = 24/7
  4. If cosec x = 25/15 , sin x = 15/25
  5. If sin A = 3/5 and cos A = 4/5, then, tan A = 3/4
  6. If cot A = 8/15 and sin A = 15/17, then, cos A = 8/17

  1. Solve:

1.Given tan A = 3/4, find the value of sin A and cos A.

Solution:

We know tan A = sin A/cos A = (opposite side/hypotenuse)/ (adjacent side/hypotenuse)

= opposite side/adjacent side

Sin A = opposite side/hypotenuse

Cos A = adjacent side/hypotenuse

Trigonometry Exercise 13.1In right angled triangle ABC, ∟C = 90˚, A = θ

By Pythagoras theorem,

AB2 = BC2 + CA2

= 32 + 42

= 9 + 16

= 25

AB = 5

sin A =  opposite side/hypotenuse  = 3/5

Cos A = adjacent side/hypotenuse = 4/5


  1. Given cot θ = 20/21­, determine cos θ and cosec θ

Solution:

We know that, cot θ = 20/21­

cot θ = cos θ/sin θ = (adjacent side/hypotenuse)/( opposite side/hypotenuse) = adjacent side/opposite side

Trigonometry Exercise 13.1In right angled triangle ABC, ∟C = 90˚, A = θ

By Pythagoras theorem,

AB2 = BC2 + CA2

= 202 + 212

= 400 + 441

= 841

AB = 29

sin A =  opposite side/hypotenuse  = 21/29

Cos A = adjacent side/hypotenuse = 20/29


  1. Given tan A = 7/24­, find the other trigonometric ratios of angle A.

Solution:

We know tan A = sin A/cos A = (opposite side/hypotenuse)/ (adjacent side/hypotenuse)

= opposite side/adjacent side = 7/24

Sin A = opposite side/hypotenuse

Cos A = adjacent side/hypotenuse

Trigonometry Exercise 13.1In right angled triangle ABC, ∟C = 90˚, A = θ

By Pythagoras theorem,

AB2 = BC2 + CA2

= 72 + 242

= 49 + 576

= 625

AB = 25

sin A =  opposite side/hypotenuse  = 7/25

Cos A = adjacent side/hypotenuse = 24/25

tan A = 7/24

cot A = 24/7

sec A = 25/24

cosec A = 25/7


  1. If 2 sin θ = √3, find cos θ, tan θ and cot θ + cosec θ

Solution:

sin θ = opposite side/hypotenuse  =  √3/2

Trigonometry Exercise 13.1In right angled triangle ABC, ∟C = 90˚, A = θ

By Pythagoras theorem,

AB2 = BC2 + CA2

22 = (√3)2 + CA2

4 = 3 + CA2

4 – 3 = CA2

CA = 1

cos θ = 1/2

tan θ = √3/1 = √3

cot θ + cosec θ = 1/√3 + 2/√3 = 1+2/√3 = 3/√3


  1. If 3 tan θ = 1, find sin θ, cos θ and cot θ

Solution:

3 tan θ = 1

tan θ = 1/3 = sin θ/cos θ

Trigonometry Exercise 13.1In right angled triangle ABC, we have ∟C = 90˚ and A = θ

AB2 = BC2 + CA2

AB2 = (1)2 + 32

= 1 + 9

= 10

AB = √10

Therefore, sin θ = 1/√10

cos θ = 3/√10

cot θ = 3/1 = 3


  1. If sec x = 2, then, find sin x, tan x, cot x and cot x + cosec x

Solution:

Given, sec x = 2 = hypotenuse/adjacent side = 2/1

Trigonometry Exercise 13.1In right angled triangle ABC, we have ∟C = 90˚ and A = θ

AB2 = BC2 + CA2

22 = BC2 + 12

4 – 1 = BC2

3 = BC2

BC = √3

sin x = √3/2

tan x = √3/1 = √3

cot x = 1/√3

cot x + cosec x = 1/√3 + 2/√3 = 1+2/√3 = 3/√3


  1. If 4 sin A – 3 cos A = 0, find sin A, cos A, sec A and cosec A

Solution:

4 sin A – 3 cos A = 0

sinA/cos A = 3/4

tan A = 3/4 = opposite side/adjacent side

Trigonometry Exercise 13.1In right angled triangle ABC, we have ∟C = 90˚ and A = θ

AB2 = BC2 + CA2

AB2 = 32 + 42

= 9 + 16

= 25

AB = 5 = hypotenuse

sin A = 3/5

cos A = 4/5

sec A = 5/4

cosec A = 5/3


  1. If 13 sin A = 5 and A is acute, find the value of 5sin A – 2 cos A/tan A

Solution:

13 sin A = 5

sin A = 5/13 = opposite side/hypotenuse

Trigonometry Exercise 13.1In right angled triangle ABC, we have ∟C = 90˚ and A = acute angle

AB2 = BC2 + CA2

132 = 52 + CA2

169 – 25 = CA2

144 = CA2

12 = CA = adjacent side

We have to find , 5sin A – 2 cos A/tan A

sin A = 5/13

cos A = 12/13

tan A = 5/12

5sin A – 2 cos A/tan A = (5. 5/13 – 2 . 12/13)/(5/12)

5sin A – 2 cos A/tan A = (25/13 24/13)/(5/12)

5sin A – 2 cos A/tan A  =( 1/13)/(5/12)

5sin A – 2 cos A/tan A = 12/65

 


  1. If cos θ = 5/13 and θ is acute, find the value of 5tanθ + 12cot θ/5tanθ – 12cot θ

Solution:

cos θ = 5/13 = Adjacent side/hypotenuse

Trigonometry Exercise 13.1In right angled triangle ABC, we have ∟C = 90˚ and A = θ = acute angle

AB2 = BC2 + CA2

132 = BC2 + 52

169 – 25 =BC2

144 = BC2

12 = BC = opposite side

We have to find, 5tanθ + 12cot θ/5tanθ – 12cot θ

tan θ = 12/5

cot θ = 5/12

5tanθ + 12cot θ/5tanθ – 12cot θ = (5.12/5 + 12.5/12)/(5.12/5 – 12.5/12)

= (12 + 5)/(12 – 5)

= 17/5


  1. If 13 cos θ – 5 = 0, find sin θ + cos θ/sin θ – cos θ

Solution:

13 cos θ – 5 = 0

cos θ = 5/13  = adjacent side/hypotenuse

Trigonometry Exercise 13.1In right angled triangle ABC, we have ∟C = 90˚ and A = θ = acute angle

AB2 = BC2 + CA2

132 = BC2 + 52

169 – 25 =BC2

144 = BC2

12 = BC = opposite side

Therefore, sin θ = 12/13

We have to find,  sin θ + cos θ/sin θ – cos θ

sin θ + cos θ/sin θ – cos θ = (12/13 + 5/13)/(12/135/13)

= (17/13)/(7/13)

= 17/7


 

Pythagoras Theorem Exercise 12.2 – Class 10

Pythagoras Theorem – Exercise 12.2 – Questions:

  1. Verify whether the following measures represent the sides of a right angled triangle.

(i)1, 2, √3

(ii) √2, √3, √5

(iii) 6√3,  12, 6

(iv) m2  – n2, 2mn, m2 + n2

  1. Pythagoras Theorem Exercise 12.2In triangle ∆ABC, a + b = 18 units, b + c = 25 units and c + a = 17 units. What type of triangle is ABC? Give reason.

 

 

3. In ∆ABC CD⊥AB, CA = 2AD and BD = 3AD. Prove that ∠BCA = 90˚

4. Pythagoras Theorem Exercise 12.2The shortest distance AP from a point A to AP is 1 cm. Q and R are respectively 15cm and  20 cm from A and on  opposite  sides  of AP. Prove that ∠QAR = 90˚

 

Pythagoras Theorem Exercise 12.25. In the isosceles ∆ABC , AB = AC, BC = 18 cm , AD⊥BC, AD = 12 cm , BC is produced to E and AE = 20cm. Prove that ∠BAE = 90˚

 

 

 

Pythagoras Theorem Exercise 12.26.In the quadrilateral ABCD, ∠ADC  = 90˚, AB = 9 cm, BC = AD  = 6 cm and CD = 3 cm. Prove that ∠ACB = 90˚

 

 

 

  1. Pythagoras Theorem Exercise 12.2ABCD is a rectangle. P is any point outside it such that PA2 + PC2 = BA2 + AD2. Prove that ∠APC = 90˚

 

 

 


Pythagoras Theorem – Exercise 12.2 – Solutions:

  1. Verify whether the following measures represent the sides of a right angled triangle.

(i)1, 2, √3

(ii) √2, √3, √5

(iii) 6√3,  12, 6

(iv) m2  – n2, 2mn, m2 + n2

Solution:

(i)1, 2, √3

Sides are: 1, 2, √3

Squares of the sides are: 12, 22, (√3)2

i.e., 1, 4, 3

Sum of areas of squares on the two smaller sides: 1 + 3 = 4

Therefore,  square on the longest side of the triangle is equal to the sum of squares on the other two sides.

By converse Pythagoras theorem, those two smaller sides must contain a right angle.

Hence the sides 1, 2 and √3 form the sides of a right angled triangle with hypotenuse √3 units , 1 and 3 units as the sides containing the right angle.

 

(ii) √2, √3, √5

Sides are: √2, √3, √5

Squares of the sides are: (√2)2, (√3)2, (√5)2

i.e., 2, 3, 5

Sum of areas of squares on the two smaller sides: 2 + 3 = 5

Therefore,  square on the longest side of the triangle is equal to the sum of squares on the other two sides.

By converse Pythagoras theorem, those two smaller sides must contain a right angle.

Hence the sides √2, √3 and √5 form the sides of a right angled triangle with hypotenuse √5 units , √2 and √3 units as the sides containing the right angle.

 

(iii) 6√3,  12, 6

Sides are: 6√3,  12, 6

Squares of the sides are: (6√3)2, 122, 62

i.e., 108, 144, 36

Sum of areas of squares on the two smaller sides: 108 + 36 = 144

Therefore,  square on the longest side of the triangle is equal to the sum of squares on the other two sides.

By converse Pythagoras theorem, those two smaller sides must contain a right angle.

Hence the sides 6√3, 12 and 6 form the sides of a right angled triangle with hypotenuse 12 units , 6√3 and 6 units as the sides containing the right angle.

 

(iv) m2  – n2, 2mn, m2 + n2

Sides are: m2  – n2, 2mn, m2 + n2

Squares of the sides are: (m2  – n2)2, (2mn)2, (m2 + n2)2

i.e., m4 – 2m2n2 + n2, 4m2n2, m4 + 2m2n2 + n2

Sum of areas of squares on the two smaller sides: m4 – 2m2n2 + n2 + m4 + 2m2n2 + n2 = 4m2n2

Therefore,  square on the longest side of the triangle is equal to the sum of squares on the other two sides.

By converse Pythagoras theorem, those two smaller sides must contain a right angle.

Hence the sides m2  – n2, 2mn and m2 + n2 form the sides of a right angled triangle with hypotenuse 2mn units , m2  – n2 and m2 + n2 units as the sides containing the right angle.


  1. Pythagoras Theorem Exercise 12.2In triangle ∆ABC, a + b = 18 units, b + c = 25 units and c + a = 17 units. What type of triangle is ABC? Give reason.

Solution:

a + b = 18————–(1)

b + c = 25————-(2)

c + a = 17————-(3)

Adding (1), (2) and (3), we get,

a + b + b + c + c + a = 17 + 25 + 18

2a + 2b + 2c = 60

a + b + c = 30 ————(*)

From (1), we have

18 + c = 30

c = 30 – 18 = 12

From (2) in (*) we have,

a + 25 = 30

a = 30 – 25

a = 5

Substitute the value of a and c in (*)

5 + b + 12 = 30

b = 30 – 12 – 5 = 13

Now, square on the longest side of the triangle is equal to the sum of squares on the other two sides.

a2 + c2 = b2

52 + 122 = 132

25 + 144 = 169

169 = 169


  1. Pythagoras Theorem Exercise 12.2In ∆ABC CD⊥AB, CA = 2AD and BD = 3AD. Prove that ∠BCA = 90˚

Solution:

Given, in ∆ABC CD⊥AB, CA = 2AD and BD = 3AD

We need to prove ∠BCA = 90˚

In ∆CDA, by Pythagoras theorem,

CA2 = CD2 + DA2

(2m)2 = (CD)2 + m2

4m2 = CD2 + m2

4m2 – m2 = CD2

3m2 = CD2

CD = √3m

In ∆CBD, by Pythagoras Theorem,

BC2 = BD2 + CD2

BC2 = (3m)2 + (√3m)2

BC2 = 9m2 + 3m2

BC2 = 12m2

BC = √12.m

In ∆BCA, by Pythagoras Theorem,

BA2 = BC2 + AC2

(4m)2 = (√12m)2 + (2m)2

16m2 = 12m2 + 4m2

16m2 = 16m2

LHS = RHS

Thus, ∠BCA = 90˚


  1. Pythagoras Theorem Exercise 12.2The shortest distance AP from a point A to AP is 1 cm. Q and R are respectively 15cm and  20 cm from A and on  opposite  sides  of AP. Prove that ∠QAR = 90˚

Solution:

AP = 12 cm

AQ = 15 cm

AR  = 20 cm

In ∆QAR,

QR2 = QA2  + AR2

= 152 + 202

= 225 + 400

= 625

Therefore, QR = 25.

In  triangle QAR, by Pythagoras theorem,

QR2 = QA2 + RA2

252 = 152 + 202

625 =  225 + 400

625 = 625


  1. In the isosceles ∆ABC , AB = AC, BC = 18 cm , AD⊥BC, AD = 12 cm , BC is produced to E and AE = 20cm. Prove that ∠BAE = 90˚

Solution:

Given, ∆ABC , AB = AC, BC = 18 cm , AD⊥BC, AD = 12 cm , BC is produced to E and AE = 20cm.

To prove ∠BAE = 90˚

Proof:

In triangle ADE, by Pythagoras Theorem,

AE2 = AD2 + DE2

202 = 122 + DE2

DE2 = 202 – 122

DE2 = 400 – 144

DE2 = 256

DE = 16 cm

In triangle ADB, by Pythagoras Theorem,

AB2 = BD2 + DA2

AB2 = 92 + 122

AB2 = 81 + 144

AB2 = 225

AB = 15 cm

In triangle ABE, by Pythagoras theorem,

BE2 = AB2 + AE2

252 = 152 + 202

625 = 225 + 400

625 = 625

Therefore, ∠BAE =  90˚


Pythagoras Theorem Exercise 12.26.In the quadrilateral ABCD, ∠ADC  = 90˚, AB = 9 cm, BC = AD  = 6 cm and CD = 3 cm. Prove that ∠ACB = 90˚

Solution:

In triangle ADC, Pythagoras theorem,

AC2 = AD2 + DC2

AC2 = 62 + 32

= 36 + 9

= 45

AC = √45

In triangle ABC ,by  Pythagoras theorem,

AB2 = BC2 + AC2

92 = 45 + 36

81 = 81

Therefore, ∠ACB =  90˚


  1. Pythagoras Theorem Exercise 12.2ABCD is a rectangle. P is any point outside it such that PA2 + PC2 = BA2 + AD2. Prove that ∠APC = 90˚

Solution:

Given, ABCD is a rectangle. P is any point outside it such that PA2 + PC2 = BA2 + AD2

Construction: Join  AC

Proof:

In triangle ABC

AC2 = AB2 + BC2

AC2 = AB2 + AD2 (BC = AD) …….(1)

Given:

(PA)2+ PC2 = BA2 + AD2 ………….(2)

From (1) and (2),

AC2 = PA2 + PC2

Therefore, ∠APC =  90˚


 

Pythagoras Theorem Exercise 12.1 – Class 10

Pythagoras Theorem – Exercise 12.1 – Questions:

a. Numerical problems based on Pythagoras theorem.

  1. The sides of a right angled triangle containing the right angle are 5 cm and 12 cm, find its hypotenuse.
  2. Find the length often diagonal of a square of side 12 cm.
  3. The length of the diagonal of a rectangular playground is 125m and the length of one side is 75m. find the length of the other side,
  4. In triangle LAW, LAW = 90, LNA = 90 , LW = 26cm, LN = 6 cm and AN = 8 cm. Calculate the length of WA.
  5. A door of width 6 meter has an arch above it having a height of 2m. Find the radius of the arch.
  6. A peacock on a pillar of 9 feet height on seeing a snake coming towards its hole situated just below the pillar will fly tp catch it. Of both poses the same speed, how far from the pillar they are going to meet?

b. Riders based on Pythagoras theorem.

  1. In triangle MGN, MP⊥GN. If MG = a units, MN = b  units, GP = c units and PN = d units. Prove that (a + b)(a – b) = (c + d)(c – d)
  2. In right angled triangle ABC, ABC = 90, BD ⊥ AC. If AB = c units, BC  = a  units, BD  = p units, CA = b  units. Prove that

1/a^2 + 1/c^2 = 1/p^2

  1. Derive the formula for height and area of an equilateral triangle.

Pythagoras Theorem – Exercise 12.1 – Solution:

a. Numerical problems based on Pythagoras theorem.

  1. The sides of a right angled triangle containing the right angle are 5 cm and 12 cm, find its hypotenuse.

Solution:

Pythagoras Theorem Exercise 12.1In triangle ABC,

AC2 = AB2 + BC2

= 122 + 52

= 144 + 25

= 169

AC = √169 = 13 cm

 

  1. Find the length often diagonal of a square of side 12 cm.

Solution:

Pythagoras Theorem Exercise 12.1By Baudhayana Theorem to squares,

square of the diagonal = 2 x square of length of its sides

AC2 = 2 x AB2

= 2 x 122

AC = √(2×122) = 12√2 cm

 

  1. The length of the diagonal of a rectangular playground is 125m and the length of one side is 75m. find the length of the other side.

Solution:

Pythagoras Theorem Exercise 12.1BD = 125m and AD = 75 m

In triangle ABD, by pythagoras theorem,

BD2 = AD2 + AB2

1252 =  752 + AB2

15625 = 5625 + AB2

15625 – 5625 = AB2

10000 = AB2

AB = 100 m

 

  1. In triangle LAW, LAW = 90, LNA = 90 , LW = 26cm, LN = 6 cm and AN = 8 cm. Calculate the length of WA.

Solution:

Pythagoras Theorem Exercise 12.1Given LW = 26 cm, NA  = 8 cm and LN = 6 cm

In triangle LNA, by Pythagoras theorem,

LA2 = LN2 + NA2

LA2 = 36 + 64

= 100

LA = 10 cm

In triangle LAW, by Pythagoras theorem

262 = 102 + WA2

676 = 100 + WA2

576 = WA2

WA = √576 = 24 cm

 

  1. A door of width 6 meter has an arch above it having a height of 2m. Find the radius of the arch.

Solution:

Pythagoras Theorem Exercise 12.1LetQ be the centre o the arch. join AQ and PQ.

Let PQ = x cm and AQ = QB = x + 2

In triangle QAP, by Pythagoras theorem,

AP2 = AQ2 + PQ2

(x + 2)2 = x2 + 32

x2 + 4x + 4 = x2 + 9

4x + 4 = 9

4x = 9 – 4 = 5

x = 5/4 = 1.25 cm

Therefore, radius of the arch = x + 2 = 1.25 + 2 = 3.25 cm

 

  1. A peacock on a pillar of 9 feet height on seeing a snake coming towards its hole situated just below the pillar will fly to catch it. Of both poses the same speed, how far from the pillar they are going to meet?

Solution:

Pythagoras Theorem Exercise 12.1Given, height of the pillar PQ = 9 feet and distance of QS = 27 feet

Therefore, RQ = 27 – x

In triangle PQR, by Pythagoras theorem,

PR2 = PQ2 + QR2

x2 = 92 + (27 – x)2

x2 = 81 + 729 – 54x + x2

54x = 81 + 729 = 810

x = 810/54 = 15 feet

Therefore QR = 27 – x = 27 – 15 = 12 feet.

 

b.Riders based on Pythagoras theorem.

  1. In triangle MGN, MP⊥GN. If MG = a units, MN = b  units, GP = c units and PN = d units. Prove that (a + b)(a – b) = (c + d)(c – d)

Solution:

Pythagoras Theorem Exercise 12.1Given, in triangle MGN, MP⊥GN. If  MG  = a units, MN = b  units, GP = c units and PN = d units.

In  triangle MPG, by Pythagoras theorem,

MG = MP  + GP

a = MP + c ————(1)

In  triangle MPN, by Pythagoras theorem,

MN = MP +  PN

b = MP + d —————(2)

subtract (1) from (2),

a2 – b2 = (MP + c)2 – (MP + d)2

a2 – b2 =  MP2 – MP2 + c2 – d2

a2 – b2 = c2 – d2

(a + b)(a – b) =  (c + d)(c – d)

 

  1. In right angled triangle ABC, ABC = 90, BD ⊥ AC. If AB = c units, BC  = a  units, BD  = p units, CA = b  units. Prove that

1/a^2 + 1/c^2 = 1/p^2

Solution:

Pythagoras Theorem Exercise 12.1Given, in right angled triangle ABC, ABC = 90, BD ⊥  AC. If AB  = c units, BC  = a  units, BD  = p units, CA = b  units.

Proof:

BD2 = AD x DC

p2 = a x DC

BC2 = AC x DC

a2 = b x DC

AB2 = AC x AD

c2 = b x AD

Now, 1/a^2 + 1/c^2 = 1/p^2

= 1/b [ 1/DC + 1/AD]

= 1/b [ AD + DC/DC.AD]

= 1/bAC/p^2]

= 1/b [b/p^2]

= 1/p^2

 

  1. Derive the formula for height and area of an equilateral triangle.

Solution:

Pythagoras Theorem Exercise 12.1AB=AC=a, BP=PC=a2, AP=h,

AP perpendicular to BC.

PROOF:

In triangle APC, by Pythagoras Theorem,

AC= AP2 + AC2

a = h2 + (a2 ) 2

a2 = h2 + (a4) 2

a 2– (a4 )2 = h2

4a2a^24 = h2

3a^24 = h

h = √(3𝑎^24) = a/2√3

Now area of triangle = 1/2 x b x h

= 1/2 x a x a/2 √3

= √3 a/4

Circles Exercise 10.8 Solution – Class 10

<script async src=”//pagead2.googlesyndication.com/pagead/js/adsbygoogle.js”></script>
<script>
(adsbygoogle = window.adsbygoogle || []).push({
google_ad_client: “ca-pub-7764692215515803”,
enable_page_level_ads: true
});
</script>

Circles Exercise 10.8 – Questions:

I(A).

  1. Draw two congruent circles of radii 3 cm, having their centres 10 cm apart, Draw a direct common tangent
  2. Draw two direct common tangents to two congruent circles of radii 3.5 cm and whose distance between them is 3 cm
  3. Draw two direct common tangent to two externally toucan circles of radii 4.5 cm
  4. Draw a pair of direct common tangents to two circles of radii 2.5 cm whose centres are at 4 cm apart.

(B)

  1. Construct a direct common tangent to two circles of radii 5 cm and 2 cm whose centres are 3 cm apart.
  2. Draw a direct common tangent to two internally touching circles of radii 4.5 cm and 2.5 cm
  3. Construct a direct common tangent to two internally touching circles of radii 4cm and 2 cm whose centres are 8 cm apart. Measure and verify the length of the tangent.
  4. Two circles of radii 5.5 cm and 3.5 cm touch each other externally. Draw a direct common tangent and measure its length.
  5. Draw direct common tangents to two circles of radii 5 cm and 3 cm having their centres 5 cm apart.
  6. Two circles of radii 6 cm and 3 cm at a distance of 1 cm, Draw a direct common tangent, measure and verify its length.

Circles Exercise 10.8 – Solution:

I(A).

  1. Draw two congruent circles of radii 3 cm, having their centres 10 cm apart, Draw a direct common tangent

Solution:

Circles Exercise 10.8 Solution

  1. Draw two direct common tangents to two congruent circles of radii 3.5 cm and whose distance between them is 3 cm

Solution:

Circles Exercise 10.8 Solution

  1. Draw two direct common tangent to two externally toucan circles of radii 4.5 cm

Solution:

Circles Exercise 10.8 Solution

  1. Draw a pair of direct common tangents to two circles of radii 2.5 cm whose centres are at 4 cm apart.

Solution:

Circles Exercise 10.8 Solution

 (B)

  1. Construct a direct common tangent to two circles of radii 5 cm and 2 cm whose centres are 3 cm apart.

Solution:

Circles Exercise 10.8 Solution

  1. Draw a direct common tangent to two internally touching circles of radii 4.5 cm and 2.5 cm

Solution:

Circles Exercise 10.8 Solution

  1. Construct a direct common tangent to two internally touching circles of radii 4cm and 2 cm whose centres are 8 cm apart. Measure and verify the length of the tangent.

Solution:

Circles Exercise 10.8 Solution

  1. Two circles of radii 5.5 cm and 3.5 cm touch each other externally. Draw a direct common tangent and measure its length.

Solution:

Circles Exercise 10.8 Solution

  1. Draw direct common tangents to two circles of radii 5 cm and 3 cm having their centres 5 cm apart.

Solution:

Circles Exercise 10.8 Solution

  1. Two circles of radii 6 cm and 3 cm at a distance of 1 cm, Draw a direct common tangent, measure and verify its length.

Solution:

Circles Exercise 10.8 Solution

Circles Exercise 10.7 Solution – Class 10

Circles Exercise 10.7 – Question

  1. Draw two circles of radii 5 cm and 2 c touching externally.
  2. Construct two circles of radii 4.5 cm and 2.5 cm whose centres are at 7 cm apart.
  3. Draw two circles of radii 4 cm and 2.5 cm touching internally. Measure and verify the distance between their centres.
  4. Distance between the centres of two circles touching internally is 2 cm. If the radius of one of the circles is 4.8 cm, find the radius of the other side and hence draw the touching circles.

Circles Exercise 10.7 – Solution

  1. Draw two circles of radii 5 cm and 2 c touching externally.

Solution:

Circles Exercise 10.7

  1. Construct two circles of radii 4.5 cm and 2.5 cm whose centres are at 7 cm apart.

Solution:

Circles Exercise 10.7

  1. Draw two circles of radii 4 cm and 2.5 cm touching internally. Measure and verify the distance between their centres.

Solution:

Circles Exercise 10.7

  1. Distance between the centres of two circles touching internally is 2 cm. If the radius of one of the circles is 4.8 cm, find the radius of the other side and hence draw the touching circles.

Solution:

Circles Exercise 10.7

Circles Exercise 10.6 Solution – Chapter Circles – Class 10

Circles Exercise 10.6 – Questions:

I. Numerical problems on touching circles.

  1. Three circles touch each other externally. Find the radii of the circles if the sides of the triangle formed by joining the centres are 7 cm, 8 cm and 9 cm respectively.
  2. Three circles with centres A, B and C touch each other as shown in the figure. IF the radii of these circles are 8 cm, 3 cm and 2 cm respectively, find the perimeter of ∆ABC.
  3. In the figure AB = 10 cm, AC = 6 cm and the radius of the smaller circle is x cm. Find x.

II. Riders based on touching circles.

  1. A straight line drawn through the point of contact of two circles with centres A and B intersect the circles at P and Q respectively. Show that AP and BQ are parallel.
  2. Two circles with centres X and Y touch each other externally at P. Two diameters AB and CD are drawn one in each circle parallel to other. Prove that B, P and C are collinear.
  3. In circle with centre O, diameter AB and a chord AD are drawn. Another circle is drawn with OA as diameter to cut AD at C. Prove that BD = 2.OC

Circles Exercise 10.6 solution:

  1. Numerical problems on touching circles.

Circles Exercise 10.61.Three circles touch each other externally. Find the radii of the circles if the sides of the triangle formed by joining the centres are 7 cm, 8 cm and 9 cm respectively.

Solution:

Let radius of  the circles be AP = x ,  BQ = y and CR = z

AB = AP + BP = x + y = 7 cm

BC = BQ + CQ = y + z = 8 cm

AC = CR + AR = z + x = 9 cm

The perimeter of ∆ABC , AB + BC + CA = 7 + 8 + 9 = 24

AP + BP + BQ + CQ + CR + AR = 24

2x + 2y + 2z 24

x + y + z = 12

7 + z = 12⇒ z = 12 – 7 = 5 cm

x + 8 = 12 ⇒ x = 12 – 8 = 4 cm

y + 9 = 12 ⇒ y = 12 – 9 = 3 cm


  1. Circles Exercise 10.6Three circles with centres A, B and C touch each other as shown in the figure. IF the radii of these circles are 8 cm, 3 cm and 2 cm respectively, find the perimeter of ABC.

Solution:

The perimeter of ∆ABC = AB + BC + AC

AB = AM – BM = 8 – 3 = 5 cm

BC = BQ + CQ = 3 + 2 = 5 cm

AC = AN – CN = 8 – 2 = 6 cm

The perimeter of ∆ABC = AB + BC + AC = 5 + 5 + 6 = 16 cm


  1. Circles Exercise 10.6In the figure AB = 10 cm, AC = 6 cm and the radius of the smaller circle is x cm. Find x.

Solution:

In ∆OPC, ∠PCO = 90˚

PC2 = OP2 – OC2

x2 = (OQ – PQ)2 – (AC – OA)2

[Since OP = OQ – PQ, OC = AC – AO]

x2 = (5 – x)2+(6 – 5)2 [Since OQ = OA = 5]

x2 = 25 – 10x + x2 – 1

10x = 24

x = 2.4 cm


  1. Riders based on touching circles.

Circles Exercise 10.61.A straight line drawn through the point of contact of two circles with centres A and B intersect the circles at P and Q respectively. Show that AP and BQ are parallel.

Solution:

∠AOP = ∠BOQ [Vertically opposite angles]

∠APO = ∠AOP [ AO = AP radius of the circle]

∠BQO = ∠BOQ

∠APO = ∠BQO [alternate angles]

Therefore, AP||BQ


  1. Circles Exercise 10.6Two circles with centres X and Y touch each other externally at P. Two diameters AB and CD are drawn one in each circle parallel to other. Prove that B, P and C are collinear.

Solution:

∠BXP = ∠PYC [Alternate angles AB||CD]

∠BPX = ∠PBX [ XB = XP radii]

∠BPX + ∠PBX + ∠BXP = 180˚ ………….(1)

∠CPY = ∠PCY [YP = YC radii]

∠CPY + ∠PCY + ∠PYC = 180˚

2∠CPY + ∠PCY = 180˚…………(2)

From (1) and (2),

2∠BPX + ∠BXP = 2∠CPY +  ∠PYC

2∠BPX = 2∠CPY

∠BPX = ∠CPY


  1. Circles Exercise 10.6In circle with centre O, diameter AB and a chord AD are drawn. Another circle is drawn with OA as diameter to cut AD at C. Prove that BD = 2.OC

Solution:

∠ADB = 90˚

∠ACO = 90˚

In ∆ADB and ∆AOC,

∠ADB = ∠ACO = 90˚

∠A = ∠A

∆ADB ∼ ∆AOC

BD/OC = AB/AO

BD/OC = 2. AO/AO [AB = 2.OA]

BD/OC = 2

BD = 2.OC


  1. Circles Exercise 10.6In the given figure AB = 8 cm, M is the midpoint of AB. A circle with centre O touches all three semicircles as shown. Prove that the radius of this circle is shown. Prove that the radius of this circle is 1/6 AB

Solution:

In ∆OPC, ∠POC = 90˚

OC2 = OM2 + MC2 [By Pythagoras Theorem]

(CP + OP)2 = (MR – OR)2 + MC2

(2 + x)2 = (4 – x)2 + 22

4 + 4x + x2 = 16 – 8x + x2 + 4

4 + 4x = 16 – 8x + 4

12x = 16

x = 16/12  = 8/6

x = 1/6 x 8

x = 1/6 x AB [AB = 8 cm]

Circles Exercise 10.5 – Class 10

Circles Exercise 10.5 – Questions:

  1. Draw a circle of radius 6 cm and construct tangents to it from an external point 10 cm away from the centre. Measure and verify the length of the tangents.
  2. Construct a pair of tangents to a circle of radius 3.5 cm from a point 3.5 cm away from the circle.
  3. Construct a tangent to a circle of radius 5.5 cm from a point 3.5 cm away from it.
  4. Draw a pair of perpendicular tangents of length 5 cm to a circle.
  5. Construct tangents to two concentric circles of radii 2 cm and 4 cm from a point 8 cm away from the centre.

Circles Exercise 10.5 – Solutions:

  1. Draw a circle of radius 6 cm and construct tangents to it from an external point 10 cm away from the centre. Measure and verify the length of the tangents.

Solution:

Circles Exercise 10.5

Lent of the tangents, t = √(d2 – r2)

Length of the tangents, t = √(102 – 62) = √(100 – 36) =  √64  = 8 cm


  1. Construct a pair of tangents to a circle of radius 3.5 cm from a point 3.5 cm away from the circle.

Solution:

Circles Exercise 10.5


  1. Construct a tangent to a circle of radius 5.5 cm from a point 3.5 cm away from it.

Solution:

3


  1. Draw a pair of perpendicular tangents of length 5 cm to a circle.

Solution:

Circles Exercise 10.5


  1. Construct tangents to two concentric circles of radii 2 cm and 4 cm from a point 8 cm away from the centre.

Solution:

Circles Exercise 10.5