From the top of a 7 m high building, the angle of elevation of the top
Trigonometry (10)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 45°. Determine the height of the tower.
Answer
$$ \text{In } \triangle ABP, \tan 45^0 = \frac{7}{x} \Rightarrow x = 7 $$
$$ \text{In } \triangle BCQ, \tan 60^0 = \frac{h}{x} \Rightarrow h = \sqrt{3} x $$
$$ h = 7\sqrt{3} m $$
$$ \text{Height of tower } = PQ = 7 + h $$
$$ = 7+ 7\sqrt{3} = 7(1+\sqrt{3}) m $$
- Exam Year: 2023
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