Question

The angle of elevation of an object from a point on the level ground is a. Moving d metres on the ground towards the object, the angle of elevation is found to be b. Then the height of the object in metres is

(a) d tan α

(b) d cot β

(c) \(\frac{d}{\text{cot}\,\alpha+\text{cot}\,\beta}\)

(d) \(\frac{d}{\text{cot}\,\alpha+\text{cot}\,\beta}\)

Answer 1

(d) \(\frac{d}{\text{cot}\,\alpha+\text{cot}\,\beta}\)

Let the height of the object AB be h metres. 

Given, ∠ACB = α, ∠ADB = β, CD = d metres 

Let DB = \(x\) metres.

Then, in rt. ΔABD, 

\(\frac{h}{x}\) = tan β ⇒ \(x\) = h cot β             .....(i)

In rt. Δ ACB,

\(\frac{h}{x+d}\) = tan α ⇒ x + d = h cot α              ...(ii) 

∴ (ii) – (i) ⇒ d = h cot α – h cot β         ...(iii)

⇒ h = \(\frac{d}{\text{cot}\,\alpha+\text{cot}\,\beta}\).