Question

From the top of a vertical tower , the angles of depression of two cars , in the same straight line with the base of tower , at an instant are found to be 45 and 60 . If the cars are 100 m apart and are on the same side of the tower , find the height of the tower . [ Use $\sqrt{3}$ = 1.73 ]

Answer 1

Let OP be the tower and points A and B be the positions of the cars.

Now,

$AB=100$ $m,$ $\angle OAP={60}^o,$ $\angle OBP={45}^o$

Let $OP=h$

In $△AOP$,

$tan{60}^o=\frac{OP}{OA}$

$\sqrt{3}=\frac{h}{OA}$

$OA=\frac{h}{\sqrt{3}}$

Also, in $△BOP$,

$tan{45}^o=\frac{OP}{OB}$

$1=\frac{h}{OB}$

$OB=h$

Now,

$OB-OA=100$

$h-\frac{h}{\sqrt{3}}=100$

$\frac{h(\sqrt{3}-1)}{\sqrt{3}}=100$

$h=\frac{100\sqrt{3}}{\sqrt{3}-1}$

$h=236.6$ $m$

hence, the height of the tower is $236.6$ $m$.