Question

the angle of elevation of the top of top of a tower at a distance 150m from its foot on a horizontal plane is found to be 30 find the hight of the tower​

Answer 1

Answer:

50

$\sqrt{3}$

Step-by-step explanation:

The distance from the foot of the tower is 150m and the angle of elevation is 30 degrees.

= tan30 degree=Perpendicular/Base

= 1/root3=P/150

= 150/root3=P

Then, by rationalising the denominator the answer will be

= P=50root3

= So the height of the tower is 50root3

Hope it helps.

Answer 2

Answer:

Height of the tower is 51.96 m.

Step-by-step explanation:

Let AB be the tower and O be the point of observation.

Then OA = 90 cm, ∠OAB = 90°.

Let AB = h meters. Then from right Δ AOB, ∠AOB = 30°.

We have,

$\frac{AB}{OA}=\frac{Perpendicular}{Base}=tan30^\o$

$\frac{h}{90^0}=\frac{1}{\sqrt3}$

$h=\frac{1}{\sqrt3}\times90^0$

$h=\frac{90}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}$

$h=\frac{90}{3}\times\sqrt3$

$h=30\times\sqrt3$

$h=30\times1.732$

∴ The height of the tower is 51.96 m.