Question

A vertical tower 50 ft. high Stands on a sloping ground. The foot of the tower is at the. same level as the middle part of a vertical flag pole. From the top of the tower the angle of depression of the top and bottom of the pole are 15 and 45 respectively. Then the length of the pole is 100 $\sqrt{3}$.

• A. True
• B. False

Answer 1

The correct option is B False

$OC=50$ is the tower standing on a sloping ground $OB$

$BA=2h$ is the flag pole whose mid point $M$ is in level with $O$

The angles of depression of $A$ and $B$ as seen from $C$ ,the top of the tower , are ${15}^{\circ },{45}^{\circ }$ respectively

Let $BD=x=OM=AE$

In $△CAE,50-h=x\tan {15}^{\circ }$

In $△CBD,50+h=x\tan {45}^{\circ }$

$⟹\frac{50-h}{50+h}=\frac{\tan {15}^{\circ }}{\tan {45}^{\circ }}$

Applying compoundo and dividendo rule

$⟹\frac{2h}{100}=\frac{\tan {45}^{\circ }-\tan {15}^{\circ }}{\tan {45}^{\circ }+\tan {15}^{\circ }}=\frac{\sin {45}^{\circ }\cos {15}^{\circ }-\sin {15}^{\circ }\cos {45}^{\circ }}{\sin {45}^{\circ }\cos {15}^{\circ }+\sin {15}^{\circ }\cos {45}^{\circ }}=\frac{\sin {30}^{\circ }}{\sin {60}^{\circ }}=\frac{1}{\sqrt{3}}$

$2h=\frac{100}{\sqrt{3}}$

So the given relation is $\text{False}$