Question

From a window $60m$ high above the ground) of a house 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 ${60}^{\circ }$ and ${45}^{\circ }$ respectively. Show that the height of the opposite house is $60(1+\sqrt{3})$ metres.

Answer 1

From right bottom triangle: $\tan 45=\frac{60}{base}$

$⟹base=60$

From top triangle:$\tan 60=\frac{x}{base}$

$⟹x=60\sqrt{3}$

Therefore Height of the house = $60+60\sqrt{3}=60(1+\sqrt{3})$