From the top of a 7 m high building, the angle of elevation of the top of a cable tower is $60°$ and the angle of depression of its foot is $30°$. Determine the height of the tower.

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Step 1: Building of height 7 m with cable tower, angles of elevation 60 degrees and depression 30 degrees
Step 2: Let AC be $h$ m, $BC = DE = 7$ m, $AB = (h - 7)$ m. $\angle AEB = 60°$ and $\angle BEC = 30°$. $	herefore \angle ECD = 30°$.
Step 3: Let $CD = x$ m. $\frac{DE}{CD} = \frac{7}{x} = 	an 30° \Rightarrow x = 7\sqrt{3} \Rightarrow BE = 7\sqrt{3}$
Step 4: $\frac{AB}{BE} = 	an 60° \Rightarrow \frac{h - 7}{7\sqrt{3}} = \sqrt{3}$
Step 5: $h - 7 = 7\sqrt{3} 	imes \sqrt{3} = 21 \Rightarrow h = 28$. Height of tower $= 28$ m.

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