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The angle of elevation of the top of an unfinished tower at a distance of 75m from its base is 30°. How much higher must the tower be raised so that the angle of elevation of its top at the same point may be 60°?

Solution

✅ Correct Option: 2

Let the current height of the tower be h1h_1 and the final height be h2h_2. The distance from the base is 75m.

Using the tangent ratio in the first triangle with angle of elevation 30°:

tan⁡(30°)=h175\tan(30°) = \frac{h_1}{75}

13=h175\frac{1}{\sqrt{3}} = \frac{h_1}{75}

h1=753h_1 = \frac{75}{\sqrt{3}}

Rationalizing:

h1=753×33h_1 = \frac{75}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}

h1=7533h_1 = \frac{75\sqrt{3}}{3}

h1=253h_1 = 25\sqrt{3} m


Using the tangent ratio in the second triangle with angle of elevation 60°:

tan⁡(60°)=h275\tan(60°) = \frac{h_2}{75}

3=h275\sqrt{3} = \frac{h_2}{75}

h2=753h_2 = 75\sqrt{3} m


The additional height required is the difference between the final height and current height:

Additional height =h2−h1= h_2 - h_1

=753−253= 75\sqrt{3} - 25\sqrt{3}

=503= 50\sqrt{3} m

Therefore, the tower must be raised by 50350\sqrt{3} meters.

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