Question

The angle of elevation of the top of a tower from a point a 60 meters from the foot is 30° the hight of the tower

Answer 1

Given:-

• Distance between tower and a point is 60m.
• Angle of elevation is 30°.

To find:-

• Height of tower.

Solution:-

Let,

The Height of tower be AB.

Angle of elevation be ∠C

And Distance between point and tower be BC

So,

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

➣ AB = ?

➣ BC = 60 m

➣ tan 30° =1/√3

$\implies \sf \: \frac{AB}{60} = \frac{1}{ \sqrt{3} }$

• Cross multiply

$\implies \sf \: AB \sqrt{3} = 60 \\ \implies \sf \: AB \: = \frac{60}{ \sqrt{3} }$

• Rationalise the denominator.

$\implies \sf \: AB = \frac{60 \times \sqrt{3} }{ \sqrt{3} \times \sqrt{3} }$

$\implies \sf \: AB \: = \frac{60 \sqrt{3} }{3 }$

• 60 will cut 20 times with the table of 3.

$\implies \sf \: AB \: = 20\sqrt{3}$

We have taken AB as height of tower.

Therefore,

Height of tower is $\sf20 \sqrt{3}$m