Question

The angle of elevation of the top of a hill at the foot of tower is ${60}^{\circ }$ and the angle of elevation of top of the tower from the foot of the hill is ${30}^{\circ }$. If the tower is 50 m high, what is the height of the hill ?

Answer 1

Let h be the height of the hill and x m be the distance between the foot of the hill and foot of the tower.
In rt.$△$ ABC,
$\cot {60}^{\circ }=\frac{x}{h}x=h\cot {60}^{\circ }....\left(1\right)Inrt.△DBC,\cot {30}^{\circ }=\frac{x}{50}x=50cot{30}^{\circ }.....\left(2\right)Equating\left(1\right)and\left(2\right)h\cot {60}^{\circ }=50\cot {30}^{\circ }h=50\cot {30}^{\circ }/\cot {60}^{\circ }h=50\times \frac{\frac{\sqrt{3}}{1}}{\sqrt{3}}=50\times 3=150m$
Therefore the height of the hill is 150 m.