Question 73

From the top of a platform 7 m high, the angle of elevation of a tower was $$30^\circ$$. If the platform was positioned $$50\sqrt{3 }$$m away from the tower, how tall was the tower?

Solution

In triangle ABC BC = DE = $$50\sqrt{3}$$, BD = 7 m

$$\tan30^{\circ\ }=\frac{AC}{BC}=\frac{AC}{50\sqrt{3}}$$

$$\frac{1}{\sqrt{\ 3}}=\frac{AC}{50\sqrt{3}}$$

$$AC=50\ m$$

Length of the tower = 50 + 7 = 57 m

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