Question

From the top of a 120m high tower, a man observes two cars on the opposite sides of the tower and in straight line with the base of tower with angles of depression as ${60}^{\circ }$ and ${45}^{\circ }$. Find the distance between the cars. (Take $\sqrt{3}=1.732)$

Answer 1

$tan60=\frac{AC}{BC}$

$\sqrt{3}=\frac{120}{BC}$

$BC=120\sqrt{3}=69.28m$

Now,

$tan45=\frac{AC}{CD}$

$1=\frac{120}{CD}$

$CD=120$

Now, distance between the cars = BD
= BC + CD
= 69.28+120
= 189.28 m