Trigonometry Exercise 13.5 – Class 10

Trigonometry Exercise 13.5 – Questions:

  1. Find the value of ‘x’:

(i)

Trigonometry Exercise 13.5

(ii)

Trigonometry Exercise 13.5

(iii)

Trigonometry Exercise 13.5

(iv)

Trigonometry Exercise 13.5

(v)

Trigonometry Exercise 13.5

II.

  1. A tall building casts a shadow of 300 m long when the sun’s altitude (elevation) is 30˚. Find the height of the tower.
  2. From the top o a building 50√3 m high, the angle of depression of an object on the ground is observed to be 45˚. Find the distance of the object from the building.
  3. A tree is broken over by wind forms a right angled triangle with the ground. If the broken part makes an angle of 60˚ with the ground and the top of the tree is now 20m from its base, how tall was the tree?
  4. The angle of elevation of the top of a flag-post from a point on a horizontal ground is found to be 30˚. On walking 6 m towards the post, the elevation increased by 15˚. Find the height of the flag-post.
  5. The angles of elevation of the top of the cliff as seen from the top and bottom of a building are 45˚ and 60˚ respectively. If the height of the building is 24 m, find the height of the cliff.
  6. From the top a building 16 m high, the angular elevation of the top of a hill is 60˚ and the angular depression of the foot of the hill is 30˚. Find the height of the hill.
  7. Find the angle of depression if an observer 150 m tall looks at the tip of his shadow which is 150√3 cm from his foot.
  8. From a point 50 m above the ground the angle of elevation of a cloud is 30˚ and the angle of depression of its reflection in water is 60˚. Find the height of the cloud above the ground.

Trigonometry Exercise 13.5 – Solutions:

  1. Find the value of ‘x’:

(i)

Trigonometry Exercise 13.5

Solution:

In ∆ABC , ∠ACB = 45˚

tanθ  =  AB/BC

tan 45˚ = x/60

1 = x/60

x = 60 m

 

 (ii)

Trigonometry Exercise 13.5

Solution:

In ∆PRQ , ∠RQP = 60˚

tanθ  =  RP/PQ

tan 60˚ = 90/x

√3 = 90/x

√3 x = 90

x = 30√3 units

 

 (iii)

Trigonometry Exercise 13.5

Solution:

In ∆LKM , ∠KLM = 30˚

tanθ  =  KM/LM

tan 30˚ = x/100

1/√3 = x/100

x√3 = 100

x = 100/√3 units

 

 (iv)

Trigonometry Exercise 13.5

Solution:

In ∆XYZ , ∠XZY = 45˚

cosθ  =  YZ/XZ

cos 45˚ = x/100

1/√2 = x/100

x√2 = 100

x = 100/√2 units

 

 (v)

Trigonometry Exercise 13.5

Solution:

In ∆DEF , ∠DEF = x

tanθ  =  DF/FE

tan x = 75/75

tan x = 1

We know, tan 45˚ = 1

tan x = tan 45˚

x = 45˚

 

II.

  1. A tall building casts a shadow of 300 m long when the sun’s altitude (elevation) is 30˚. Find the height of the tower.

Solution:

In ∆ABC , ∠ACB = 30˚

tanθ  =  AB/BC

tan 30˚ = x/300

1/√3 = x/300

x√3 = 300

x = 300/√3 = 100√3

 

  1. From the top of a building 50√3 m high, the angle of depression of an object on the ground is observed to be 45˚. Find the distance of the object from the building.

Solution:

Trigonometry Exercise 13.5Let PQ be the height f the building, PQ = 50√3, QR be the distance between the building and object.

Angle of depression = 45˚

Since, PM||QR, ∠MPR = ∠PRQ (alternate angles)

∠MPR = 45˚

Therefore, ∠PRQ = 45˚

In ∆PQR , ∠MPR = 45˚

tanθ  =  PQ/QR

tan 45˚ = 50√3/QR

1 = 50√3/QR

QR = 50√3 m

 

  1. A tree is broken over by wind forms a right angled triangle with the ground. If the broken part makes an angle of 60˚ with the ground and the top of the tree is now 20m from its base, how tall was the tree?

Solution:

Trigonometry Exercise 13.5Let AC is the full length of the tree. AB is broken and  bends as BD. Therefore AB = BD.

Given, CD = 20 m and ∠BDC = 60˚

In ∆BCD , ∠BDC = 60˚

tanθ  =  BC/CD

tan 60˚ = 20/CD

√3 = BC/20

20√3 = BC

cosθ = CD/BD

1/2 = 20/BD

40 = BD

To find the full length of the tree , AC = BC + BD(since  AB = BD)

= 20√3 + 40

= 20(√3 + 2) m

 

  1. The angle of elevation of the top of a flag-post from a point on a horizontal ground is found to be 30˚. On walking 6 m towards the post, the elevation increased by 15˚. Find the height of the flag-post.

Solution:

Trigonometry Exercise 13.5Given ∠BAC = 30˚, ∠BDC = 45˚

tan30˚ = BC /AD + DC

1/√3 = BC/6+x

√3 .BC = 6 + x

BC  = 6+x/√3 ………….(1)

tan45˚ = BC/DC

1 = BC/x

BC = x…………..(2)

Substitute (2) in (1),

x = 6 + x/√3

√3.x = 6 + x

6 = √3.x – x

6 = x(√3 – 1)

x = 6/(√3 – 1)

 

  1. The angles of elevation of the top of the cliff as seen from the top and bottom of a building are 45˚ and 60˚ respectively. If the height of the building is 24 m, find the height of the cliff.

Solution:

Trigonometry Exercise 13.5Height od the building i.e., AD = BC = 24 m. Similarly, AB = DC,

In ∆MCD , ∠MDC = 60˚

tan θ = MC/DC

tan60˚ = 24+x/DC

√3 = 24+x/DC

DC.√3 = 24 + x

DC = 24+x/√3

In ∆AMB , ∠MAB = 45˚, MB = x

tan θ = MB/AB

tan45˚ = x/(24+x/√3)

1 = x√3/24+x

√3x = 24 + x

√3x – x = 24

(√3 – 1)x = 24

x = 24/√3 – 1

Height of the cliff, MC = MB + BC = (24 + 24/√3 – 1) m

 

  1. From the top a building 16 m high, the angular elevation of the top of a hill is 60˚ and the angular depression of the foot of the hill is 30˚. Find the height of the hill.

Solution:

Trigonometry Exercise 13.5In ∆DCE , ∠CED = 30˚, DE = x

tan θ = CD/DE

tan30˚ = 16/x

1/√3 = 16/x

x = 16√3

In ∆ABC , ∠BCA = 60˚, BC = DE = x

tan θ = AB/BC

tan 60˚ = AB/x

√3 = AB/16√3

16×3 = AB

48 = AB

Therefore, the height of the hill = AB + BE = 48 + 16 = 64 m

 

  1. Find the angle of depression if an observer 150 m tall looks at the tip of his shadow which is 150√3 cm from his foot.

Solution:

Trigonometry Exercise 13.5In ∆ABC , AB = 150 cm and BC = 150√3

tan θ = AB/BC = 150/150√3

tan θ = 1/√3

We know, tan30˚ = 1/√3

Therefore, tan θ = tan30˚

Hence, θ = 30˚

 

  1. From a point 50 m above the ground the angle of elevation of a cloud is 30˚ and the angle of depression of its reflection in water is 60˚. Find the height of the cloud above the ground.

Solution:

Trigonometry Exercise 13.5Let AC = h

BE = CD = 50 m

A’D = AD = 50 + h

In ∆ABC , ∠ABC = 60˚,

tan θ = AC/BC = h/BC

tan 30˚ = h/BC

1/√3 = h/BC

√3.h = BC

Now, in ∆A’CB ,

tan60˚ = CA’/BC = CD+DA’/BC

√3 = 50+50+h/√3h = 100+h/√3h

3h = 100 + h

2h = 100

h = 50 m

Thus, height of the cloud from the surface of lake = AD+A’D = 50 + 50  = 100m

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

Expansion

A quantity expressed as a sum of a series of terms i.e., an expression written as sum of its series of terms.

For example x (3y + z)= 3xy + 3z . Here this 3xy + 3z is expansion of x (3y + z).

Another example, (x + 3)(x +2) = x^2 + 2x + 3x + 6