Question

At a point, the angle of elevation of a tower is such that its tangent is $\frac{5}{12}$. On walking $240m$ nearer the tower, the tangent of the angle of elevation becomes $\frac{3}{4}$. Find the height of the tower.

Answer 1

In the figure, Let $AB$ be the tower, $C$ and $D$ be the positions of observations from where given that

$\tan \varphi =\frac{5}{12}$ ...(i)

And $\tan \varphi =\frac{3}{4}$ ...(ii)

Let $BC=xm.AB=ym$

Now in right-angled triangle $ABC$

$\tan \varphi =\frac{y}{x}$ ....(iii)

From (ii) and (iii), we get $\frac{3}{4}=\frac{y}{x}$

$\Rightarrow x=\frac{4}{3}y$ .... (iv)

Also in right-angled triangle $ABD$, we get

$\tan \varphi =\frac{y}{x+240}$ ...(v)

From (i) and (v), we get

$\frac{5}{12}=\frac{y}{x+240}\Rightarrow 12y=5x+1200$ ...(vi)

$\Rightarrow 12y=5\times \frac{4}{3}y+1200$ ( Using (iv))

$\Rightarrow 12y-\frac{20}{3}y=1200\Rightarrow \frac{36y-20y}{3}=1200$

$\Rightarrow 16y=3600\Rightarrow y=\frac{3600}{16}=225$

Hence, the height of the tower is $225$ metres.