## Maharashtra State Board Class 10 Maths Solutions Chapter 6 Trigonometry Problem Set 6

Question 1.

Choose the correct alternative answer for the following questions.

i. sin θ.cosec θ = ?

(A) 1

(B) 0

(C) \(\frac { 1 }{ 2 } \)

(D) \(\sqrt { 2 }\)

Answer:

(A)

ii. cosec 45° = ?

(A) \(\frac{1}{\sqrt{2}}\)

(B) \(\sqrt { 2 }\)

(C) \(\frac{\sqrt{3}}{2}\)

(D) \(\frac{1}{\sqrt{3}}\)

Answer:

(B)

iii. 1 + tan^{2} θ = ?

(A) cot^{2} θ

(B) cosec^{2} θ

(C) sec^{2} θ

(D) tan^{2} θ

Answer:

(C)

iv. When we see at a higher level, from the horizontal line, angle formed is ______

(A) angle of elevation.

(B) angle of depression.

(C) 0

(D) straight angle.

Answer:

(A)

Question 2.

If sin θ = \(\frac { 11 }{ 61 } \), find the value of cos θ using trigonometric identity.

Solution:

sin θ = \(\frac { 11 }{ 61 } \) … [Given]

We know that,

sin^{2} θ + cos^{2} θ = 1

…[Taking square root of both sides]

Question 3.

If tan θ = 2, find the values of other trigonometric ratios.

Solution:

tan θ = 2 …[Given]

We know that,

1 + tan^{2} θ = sec^{7} θ

∴ 1 + (2)^{7} = sec^{7} θ

∴ 1 + 4 = sec^{7} θ

∴ sec^{7} θ = 5

∴ sec θ = \(\sqrt { 5 }\) …[Taking square root of both sides]

Question 4.

If sec θ = \(\frac { 13 }{ 12 } \), find the values of other trigonometric ratios.

Solution:

sec θ = \(\frac { 13 }{ 12 } \) … [Given]

We know that,

1 + tan^{2} θ = sec^{2} θ

∴ sin θ = \(\frac { 5 }{ 13 } \), cos θ = \(\frac { 12 }{ 13 } \), tan θ = \(\frac { 5 }{ 12 } \), cot θ = \(\frac { 12 }{ 5 } \), cosec θ = \(\frac { 13 }{ 5 } \)

Question 5.

Prove the following:

i. sec θ (1 – sin θ) (sec θ + tan θ) = 1

ii. (sec θ + tan θ) (1 – sin θ) = cos θ

iii. sec^{2} θ + cosec^{2} θ = sec^{2} θ × cosec^{2} θ

iv. cot^{2} θ – tan^{2} θ = cosec^{2} θ – sec^{2} θ

v. tan^{4} θ + tan^{2} θ = sec^{4} θ – sec^{2} θ

vi. \(\frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}\) = 2 sec2 θ

vii. sec^{6} x – tan^{6} x = 1 + 3 sec^{2} x × tan^{2} x

Proof:

i. L.H.S. = sec θ (1 – sin θ) (sec θ + tan θ)

∴ sec θ (1 – sin θ) (sec θ + tan θ) = 1

ii. L.H.S. = (sec θ + tan θ) (1 – sin θ)

∴ (sec θ + tan θ) (1 – sin θ) = cos θ

iii. L.H.S. = sec^{2} θ + cosec^{2} θ

∴ sec^{2} θ + cosec^{2} θ = sec^{2} θ × cosec^{2} θ

iv. L.H.S. = cot^{2} θ – tan^{2} θ

= (cosec^{2} θ – 1) – (sec^{2} θ – 1)

[∵ tan^{2} θ = sec^{2} θ – 1,

cot^{2} θ = cosec^{2} θ – 1]

= cosec^{2} θ – 1 – sec^{2} θ + 1

cosec^{2} θ – sec^{2} θ

= R.H.S.

∴ cot^{2} θ – tan^{2} θ = cosec^{2} θ – sec^{2} θ

v. L.H.S. = tan^{4} θ + tan^{2} θ

= tan^{2} θ (tan^{2} θ + 1)

= tan^{2} θ. sec^{2} θ

…[∵ 1 + tan^{2} θ = sec^{2} θ]

= (sec^{2} θ – 1) sec^{2} θ

…[∵ tan^{2} θ = sec^{2} θ – 1]

= sec^{4} θ – sec^{2} θ

= R.H.S.

∴ tan^{4} θ + tan^{2} θ = sec^{4} θ – sec^{2} θ

vii. L.H.S. = sec^{6} x – tan^{6} x

= (sec^{2} x)^{3} – tan^{6} x

= (1 + tan^{2} x)^{3} – tan^{6} x …[∵ 1 + tan^{2} θ = sec^{2} θ]

= 1 + 3tan^{2} x + 3(tan^{2} x)^{2} + (tan^{2} x)^{3} – tan^{6} x …[∵ (a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}]

= 1 + 3 tan^{2} x (1 + tan^{2} x) + tan^{6} x – tan^{6} x

= 1 + 3 tan^{2} x sec^{2} x …[∵ 1 + tan^{2} θ = sec^{2} θ]

= R.H.S.

∴ sec^{3}x – tan^{6}x = 1 + 3sec^{2}x.tan^{2}x

x. We know that,

sin^{2} θ + cos^{2} θ = 1

∴ 1 – sin^{2} θ = cos^{2} θ

∴ (1 – sin θ) (1 + sin θ) = cos θ. cos θ

Question 6.

A boy standing at a distance of 48 metres from a building observes the top of the building and makes an angle of elevation of 30°. Find the height of the building.

Solution:

Let AB represent the height of the building and point C represent the position of the boy.

Angle of elevation = ∠ACB = 30°

BC = 48 m

In right angled ∆ABC,

tan 30° = \(\frac { AB }{ BC } \) … [By definition]

∴ The height of the building is 16\(\sqrt { 3 }\) m.

Question 7.

From the top of the lighthouse, an observer looks at a ship and finds the angle of depression to be 30°. If the height of the lighthouse is 100 metres, then find how far the ship is from the lighthouse.

Solution:

Let AB represent the height of lighthouse and point C represent the position of the ship.

Angle of depression ∠PAC 30°

AB = 100m.

Now, ray AP || seg BC

∴ ∠ACB = ∠PAC … [Alternate angles]

∴ ∠ACB = 30°

AB = 100m

In right angled ∆ABC,

tan 30° = \(\frac { AB }{ BC } \) …[By definition]

∴ \(\frac{1}{\sqrt{3}}=\frac{100}{\mathrm{BC}}\)

∴ BC = 100\(\sqrt { 3 }\)m

∴ The ship is 1oo\(\sqrt { 3 }\)m far from the lighthouse.

Question 8.

Two buildings are in front of each other on a road of width 15 metres. From the top of the first building, having a height of 12 metre, the angle of elevation of the top of the second building is 30°. What is the height of the second building?

Solution:

Let AB and CD represent the heights of the two buildings, and BD represent the width of the road.

Draw seg AM ⊥ seg CD

Angle of elevation = ∠CAM = 30°

AB = 12m

BD = 15m

In ꠸ ABDM,

∠B = ∠D = 90°

∠M 90° …[segAM ⊥ segCD]

∠A 90° …[Remaining angle of ꠸ABDM]

꠸ABDM is a rectangle …[Each angle is 90°]

∴ The height of the second building is 20.65 m.

Question 9.

A ladder on the platform of a fire brigade van can be elevated at an angle of 70° to the maximum. The length of the ladder can be extended upto 20 m. If the platform is 2 m above the ground, find the maximum height from the ground upto which the ladder can reach. (sin 70° = 0.94)

Solution:

Let AB represent the length of the ladder and AE represent the height of the platform.

Draw seg AC ⊥ seg BD.

Angle of elevation = ∠BAC = 70°

AB = 20 m

AE = 2m

In right angled ∆ABC,

sin 70° = \(\frac { BC }{ AB } \) …..[By definition]

∴ 0.94 = \(\frac { BC }{ 20 } \)

∴ BC = 0.94 × 20

= 18.80 m

In ꠸ACDE,

∠E = ∠D = 90°

∠C = 90° … [seg AC ⊥ seg BD]

∴ ∠A = 90° … [Remaining angle of ꠸ACDE]

∴ ꠸ACDE is a rectangle. … [Each angle is 90°]

∴ CD = AE = 2 m … [Opposite sides of a rectangle]

Now, BD = BC + CD … [B – C – D]

= 18.80 + 2

= 20.80 m

∴ The maximum height from the ground upto which the ladder can reach is 20.80 metres.

Question 10.

While landing at an airport, a pilot made an angle of depression of 20°. Average speed of the plane was 200 km/hr. The plane reached the ground after 54 seconds. Find the height at which the plane was when it started landing, (sin 20° = 0.342)

Solution:

Let AC represent the initial height and point A represent the initial position of the plane.

Let point B represent the position where plane lands.

Angle of depression = ∠EAB = 20°

Now, seg AE || seg BC

∴ ∠ABC = ∠EAB … [Alternate angles]

∴ ∠ABC = 20°

Speed of the plane = 200 km/hr

= 200 × \(\frac { 1000 }{ 3600 } \) m/sec

= \(\frac { 500 }{ 9 } \) m/sec

∴ Distance travelled in 54 sec = speed × time

= \(\frac { 500 }{ 9 } \) × 54

= 3000 m

∴ AB = 3000 m

In right angled ∆ABC,

sin 20° = \(\frac { AC }{ AB } \) ….[By definition]

∴ 0.342 = \(\frac { AC }{ 3000 } \)

∴ AC = 0.342 × 3000

= 1026 m

∴ The plane was at a height of 1026 m when it started landing.