Heights and Distances Practice Test 2

Let AB = h m be the height of the tower; B and C are two points such that BC = 20 m; ∠ADB = 30° and ∠ACB = 60°; BC = x m (say)
Now, from right triangle ABC,
$\mathit{x}=\mathit{h}\cot 60°=\frac{\mathit{h}}{\sqrt{3}}\mathit{m}$
Again, from right triangle ABD,
$\mathit{h}=\left(20+\mathit{x}\right)\tan 30°$ $=\left(20+\frac{\mathit{h}}{\sqrt{3}}\right)\times \frac{1}{\sqrt{3}}$
$\Rightarrow \mathit{h}-\frac{\mathit{h}}{3}=\frac{20}{\sqrt{3}}$
$\therefore \mathit{h}=10\sqrt{3}\mathit{m}$