A Tower Stands Vertically on the Ground. From a Point on the Ground, Which is 15 m away From the Foot of the Tower, the Angle of Elevation of the Top of the Tower is Found to be 30°. Find the Height of the Tower - Bzziii

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A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 30°. Find the height of the tower.

(a) `5\sqrt{3}` m
(b) `15\sqrt{3}` m
(c) 15 m
(d) `\frac{5}{\sqrt{3}}` m





(b) `15\sqrt{3}` m



Option (b)`15/\sqrt{3}` is the correct answer.

Let, the height of the tower be H meter.

So, AB = H meter

Distance of the point from the foot of the tower = 15m

Hence, CB = 15 m

Angle of elevation = `30^0`

∠ACB = `30^0`

Since, the tower is vertical to ground,

So, ∠ABC = `30^0`

Now,

tan C = `"side opposite to angle C"/"side adjacent to angle C"`

tan C = `"AB"/"CB"`

tan `30^0` = `"AB"/"CB"`

⇒ `1/\sqrt{3}` = `"AB"/"CB"`

⇒ `1/\sqrt{3}` = `"H"/"15"`

⇒ 15 = H`\sqrt{3}`

⇒ `15/\sqrt{3}` = H

∴ H = `15/\sqrt{3}` 

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