Question

the angle of depression of the top and bottom of a 9m high building from the top of a tower aree30 degree and 60 degree respectively.find the height of the tower and the distance between the buidingand the tower?

Answer 1

Given :

Hight of building on which observer is standing : 9 m

Angle of depression of top = 30°

Angle of depression of bottom = 60°

To find :

Height of another tower.

Solution :

We know that angle of depression of top and bottom of the building is 30° and 60° respectively.

Let BE=h,

where h be the height of the building whose height we have to find.

AB=9−h,

BC=ED

Distance between tower and building be ED=x

In ΔAED,

$\tan 60\degree \: = \frac{AE}{ED} \: \:$

so putting the values of AE = 9 and ED = x,

$\tan 60\degree \: = \frac{9}{x} \: \:$

putting the value of tan60° = √3

we get the value of x, (distance between both buildings)

$\sqrt{3} = \frac{9}{x} \\ x = \frac{9}{ \sqrt{3} } = \: \: 3 \sqrt{3}$

so distance between both buildings = 3√3 meters.

In ΔABC,

$\tan 30 \degree \: = \frac{AB}{ BC}$

Putting AB = 9-h and BC = DE = x = 3√3 above

$\tan 30 \degree \: = \frac{9 - h}{ 3 \sqrt{3} } \\ \frac{1}{ \sqrt{3} } = \frac{9 - h}{ 3 \sqrt{3} } \: \\ 9 - h = 3 \\ h = 6$

So h = Height of tower = 6 meters.