Question

from the top of a vertical tower the angle of depression of two car in the same straight line with the base of the tower at an instant are found to be 45 degree and 60 degree if the cars are 100m apart and are on the same side of the tower find the height of tower

Answer 1

Refer to the attachments please.

Answer 2

Answer:

The height of tower is 236.602 m

Step-by-step explanation:

Refer the attached figure

AB is the tower

The angle of depression of two car in the same straight line with the base of the tower at an instant are found to be 45 degree and 60 degree i.e. ∠ACB = 60° and ∠ADB = 45°

The cars are 100 m apart i.e. CD = 100 m

Let CB be x

So, DB = DC+CB = 100+x

In ΔABC

$Tan\theta = \frac{Perpendicular}{Base}$

$Tan 60^{\circ}= \frac{AB}{BC}$

$\sqrt{3}= \frac{AB}{x}$

$\sqrt{3}x= AB$   ---1

In ΔABD

$Tan\theta = \frac{Perpendicular}{Base}$

$Tan 45^{\circ}= \frac{AB}{BC}$

$1= \frac{AB}{BD}$

$1= \frac{AB}{100+x}$

$100+x=AB$   ------2

With 1 and 2

$100+x=\sqrt{3}x$

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

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

$136.6025=x$

So, $\sqrt{3}(136.6025)= 236.602 =AB$

Hence the height of tower is 236.602 m