Question

(the angle of elevation of the tip of a tower from a point on the ground which is 30m away from the foot the tower is 30° find the height of the tower
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Answer 1

Given:-

• Angle of elevation is 30°.
• Distance between tower and any point on ground is 30 m.

To find:-

• Height of tower.

Solution:-

Let AB be the tower of h meters.

And C is point on ground such that angle of elevation of the top A of tower AB is of 30°.

In ∆ABC,

BC = 30 m

∠C = 30°

• We have to find Perpendicular AB.

In ∆ABC ,

⇒ tan C = $\sf \frac{AB}{BC}$

• ∠C = 30°

⇒ tan 30° = $\sf \frac{AB}{BC}$

• tan 30° = 1/√3 , AB = h and BC = 30 m.

⇒ $\sf \frac{1}{\sqrt{3}}$ = $\sf \frac{h}{30}$

• Cross multiply

⇒ $\sf 30 =\sqrt{3} \times h$

⇒ $\sf \frac{30}{\sqrt{3}} = h$

Or ,

⇒ $\sf h = \frac{30}{\sqrt{3}}$

• Rationalise the denominator.

⇒ $\sf h = \frac{30 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

⇒ $\sf h = \frac{30\times \sqrt{3}}{3}$

⇒ $\sf h = 10 \sqrt{3}$

h is height of tower.

Therefore,

Height of tower is 10 $\sqrt{3}$ m.