Question

PT is a tower of height 2^x metres, P being the foot, T being the top of the tower. A and B are points on the same line with P. If AP = 2^{x + 1} m and BP = 192 m and if the angle of elevation of the tower as seen from B is double the angle of elevation of the tower as seen from A, then what is the value of x? 

(a) 6 

(b) 7 

(c) 8 

(d) 9

Answer 1

(c) 8

Given, PT = 2^x m, 

AP = 2^x+1 m, BP = 192 m, 

∠TAP = θ, ∠TBP = 2θ 

In rt. Δ TAP, tan θ = \(\frac{PT}{AP}\)

⇒ tan θ \(\frac{2^x}{2^{x+1}}\) = \(\frac{1}{2}\)              ..........(i)

In rt. Δ TBP, tan 2θ = \(\frac{PT}{BP}\) = \(\frac{2^x}{192}\)

⇒ \(\frac{2\,tan\,\theta}{1-\tan^2\,\theta}\) = \(\frac{2^x}{192}\) ⇒ \(\frac{2\times\frac{1}{2}}{1-\frac{1}{4}}\) = \(\frac{2^x}{192}\)         (from (i))

⇒ \(\frac{3}{4}\) x 192 = 2^x ⇒ 2^x = 256 ⇒ 2^x = 2⁸ ⇒ \(x\) = 8.