From a window 15m high above the ground in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are 30° and 45° respectively. The height of the opposite house is

∠ DPQ = 30°,

∠ ACP = ∠ QPC = 45°

AP = 150 metre

CD = building = h metre

QD = CD – CQ = CD – AP = (h – 15) metre

In Δ PQC,

$\tan 45^{\circ}=\frac{\mathrm{QC}}{\mathrm{PQ}} \Rightarrow 1=\frac{15}{\mathrm{PQ}}$

⇒ PQ = 15 metre

In $\Delta \mathrm{PQD}, \tan 30^{\circ}=\frac{\mathrm{QD}}{\mathrm{PQ}}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h-15}{15}$

$\Rightarrow h-15=\frac{15}{\sqrt{3}}$

$\Rightarrow h-15=5 \sqrt{3}$

$\Rightarrow h=15+5 \times 1.732$

$=23.66$ metre