Heights and Distances Practice Test 2

Let AB = hm be the height of the tower and two ships are situated at D and C respectively; such that, CD = 200 m; ∠ ADC = 30°; ∠ ACB = 45° and BC = xm (say)
Now from right triangle ABC,
$\tan 45°=\frac{\mathit{h}}{\mathit{x}}\Rightarrow 1=\frac{\mathit{h}}{\mathit{x}}$
$\therefore \mathit{x}=\mathit{h}$
Again from right triangle ABD,
$\tan 30°=\frac{\mathit{h}}{200+\mathit{x}}\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathit{h}}{200+\mathit{h}}$
$\Rightarrow \sqrt{3}\mathit{h}-\mathit{h}=200$
$\Rightarrow \mathit{h}=273$