Important for :

1

A

Let, the height of the tower = h metre

Given, BP = a metre and BQ = b metre

And, ∠APB and ∠AQB are complementary.

∴ ∠AQB = Θ and ∠APB = (90° – Θ)

In ΔABQ,

tanΘ = AB / BQ

tan Θ = h / b .....(i)

Now, in ΔABP,

tan (90º – Θ) = AB / BP

cot Θ = (h / a) .....(ii)

[

By multiplying both equations (i) and (ii)

tan Θ cot Θ = (h / b) × (h / a)

1 = (h

h = √ab

Hence, option A is correct.

2

3

A

Given, BC = 160 m and CD = 100 m

∴ BD = BC – CD = 160 – 100 = 60 m

Let, the height of the tower, AB = h metre

And, ∠ACB = Θ and ∠ADB = 2Θ

In ΔABC,

tab Θ = AB / BC = h / 100 .....(i)

Now, in ΔABD,

tan 2Θ = (AB / BD)

2tanΘ / (1 – tan

[∵ tan 2 Θ = 2tanΘ / (1 – tan

120 tan Θ = h(1 – tan

120 × (h / 160) = h {1 – (h / 160)

[From eq..... (i)]

3 / 4 = 1 – (h / 160)

(h / 160)

h / 160 = √(1 / 4) = 1 / 2

h = 80 m

Hence, option A is correct.

4

5

A

Let, the height of the pillar, AB = h metre.

When the sun's angle of elevation was 45°, then the length of shadow of the pillar is BD = x (let).

And, when the sun's angle of elevation is 30°, then the length of shadow of the pillar is BC.

When the sun changes from 45° to 30°, then the length of shadow of the pillar increases CD = 10 (given)

∴ BC = CD + BD = (10 + x) m

∴ BC = CD + BD = (10 + x) m

In ΔABD,

tan45º = AB / BD ⇒ 1 = h / x

tan45º = AB / BD ⇒ 1 = h / x

⇒ h = x ...(i)

Now, in ΔABC,

tan 30º = AB / BC ⇒ 1 / √3

= h / (x + 10)

⇒ h√3 – x = 10

⇒ h√3 – h = 10 [From eq. (i)]

⇒ h(√3 – 1) = 10

⇒ h = 10 / (√3 – 1) × {(√3 + 1) / (√3 + 1)}

= 5 (√3 + 1)m

Hence, option A is correct.

Now, in ΔABC,

tan 30º = AB / BC ⇒ 1 / √3

= h / (x + 10)

⇒ h√3 – x = 10

⇒ h√3 – h = 10 [From eq. (i)]

⇒ h(√3 – 1) = 10

⇒ h = 10 / (√3 – 1) × {(√3 + 1) / (√3 + 1)}

= 5 (√3 + 1)m

Hence, option A is correct.