Question

The angle of elevation of the top of the minar from the foot of the tower of height h is α and the tower subtends and angle β at the top of the minar. Then, the height of the minar is

(a) \(\frac{h\,cot\,(α\,-\,β)}{cot\,(α\,-\,β)-cot\,α}\)

(b) \(\frac{h\,tan\,(α\,-\,β)}{tan\,(α\,-\,β)-tan\,α}\)

(c) \(\frac{h\,cot\,(α\,-\,β)}{cot\,(α\,-\,β)+cot\,α}\)

(d) None of the above.

Answer 1

(a) \(\frac{h\,cot\,(α-β)}{cot\,(α-β)\,-\,cot\,α}\)

Let AB be the tower and PQ be the minar.

Given, 

AB = h, let PQ = x.

Draw AC || BQ meeting PQ in C

Then, 

BQ = AC = y (say)

∴ CQ = AB = h

⇒ PC = PQ – CQ = x – h

Given, 

∠PBQ = α, ∠APB = β.

Now,

∠PDC = α  (DC || BQ, corr. ∠s)

In Δ ADP,

∠PAD + ∠APD = ext. ∠PDC

⇒ ∠PAD = ext. ∠PDC – ∠APD = α – β.

In rt. Δ PCA,

tan (α – β) = \(\frac{PC}{AC}=\frac{x-h}{y}\)

⇒ y = \(\frac{x-h}{tan\,(α-β)}\) ...(i)

In rt. Δ PBQ,

tan α = \(\frac{PQ}{BQ}\) = \(\frac{x}{y}\)

⇒ y = \(\frac{x}{tan\,α}\)  ...(ii)

From (i) and (ii),

\(\frac{x-h}{tan\,(α-β)}\) = \(\frac{x}{tan\,α}\)

⇒ \(\frac{x-h}{x}\) = \(\frac{tan\,(α-β)}{tan\,α}\)

⇒ 1 - \(\frac{h}{x}\) = \(\frac{cot\,α}{cot\,(α-β)}\)

⇒ \(\frac{h}{x}\) = \(1-\frac{cot\,α}{cot\,(α-β)}\)

= \(\frac{cot\,(α-β)\,-\,cot\,α}{cot\,(α-β)}\)

⇒  x = \(\frac{h\,cot\,(α-β)}{cot\,(α-β)\,-\,cot\,α}\)