Heights and Distances Practice Test 1

Let AB = h m be the height of the tower; B and C are two points such that ∠ACB = 30°; ∠ADB = 45° and CD = x m (say)
From right triangle ABD,
$\tan 45°=\frac{\mathit{h}}{\mathit{B}\mathit{D}}\Rightarrow \mathit{B}\mathit{D}=\mathit{h}\mathrm{m};$
Again from right triangle ABC
$\mathrm{ta}30°=\frac{\mathit{h}}{\mathit{h}+\mathit{x}}\Rightarrow \mathit{h}+\mathit{x}=\sqrt{3}\mathit{h}$
∴ x = (1.73 – 1)h = 0.73h
Now, 0.73h m covered in 12 min
Hence, h m covered in $\frac{12}{0.73}\min \approx 16\min 23\sec$