Question

The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 metres towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.

Answer 1

Let PR = h meter, be the height of the tower.

The observer is standing at point Q such that, the distance between the observer and tower is QR = (20+x) m, where

QR = QS + SR = 20 + x

∠PQR = 30°

∠ PSR = θ

In ∆PQR,

Rearranging the terms,

We get 20 +x = √3h

⇒ x = √3h – 20 …eq.1

In ∆PSR,

tan θ = h/x

Since, angle of elevation increases by 15^o when the observer moves 20 m towards the tower.

We have,

θ = 30° + 15° = 45°

So,

tan 45^o = h/x

⇒ 1 = h/x

⇒ h = x

Substituting x=h in eq. 1, we get

h = √3 h – 20

⇒ √3 h – h = 20

⇒ h (√3 – 1) = 20

= 10 (√3 + 1)

Hence, the required height of the tower is 10 (√3 + 1) meter.

Answer 2

Let AB be the tower and C and D be the point of observation 

Given  CD = 20 m And ∠BCA = 30° and ∠BDA = 30 + 15 = 45°

Let height of tower is h

In triangle BAD,

\(\tan 45° = \frac{AB}{AD} \)

⇒ \(1 = \frac h{AD}\)

⇒ \(AD = h\)

In triangle BAC

\(\tan 30° = \frac{AB}{AC} (AC = CD + AD)\)

⇒ \(\frac 1{\sqrt 3} = \frac h{20 + h}\)

⇒ \(\sqrt 3h = 20 + h\)

⇒ \(\sqrt 3 - h = 20\)

⇒ \(h (1.732 - 1) = 20\)

⇒ \(h = \frac{20}{0.732} = 27.3\)