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A tower stands vertically on the ground. From a point on the ground $15\mathrm{m}$ away from its foot, the angle of elevation of the top is $60^{\circ}$. What is the height of the tower? ![](https://qallery.app/diagrams/v2_834a41492565/img-005.jpeg)

A$5\sqrt{3}\mathrm{m}$
B$10\sqrt{3}\mathrm{m}$
C$15\sqrt{3}\mathrm{m}$
D$30\sqrt{3}\mathrm{m}$
Answer & Solution
Correct answer: C. $15\sqrt{3}\mathrm{m}$
Form a right triangle with height as the opposite side and $15\mathrm{m}$ as the adjacent side. Then $\tan 60^{\circ}=\frac{h}{15}=\sqrt{3}$. So $h=15\sqrt{3}\mathrm{m}$.
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