A vertical tower of height H stands on the ground. From point A on the ground, the angle of elevation to the top is 45°. From point B, located 15 meters away from A (directly along the line extending from the base), the angle of elevation is 30°. What is the height of the tower?

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Accepted Answer

1. Let the distance from the tower base to A be $x$. From A: $	an 45^\circ=1=H/x$, so $H=x$.
2. B is 15 m farther, so base to B is $x+15$. From B: $	an 30^\circ=\frac{1}{\sqrt{3}}=H/(x+15)$, so $H=(x+15)/\sqrt{3}$.
3. Set $H=x$. Solve $x=(x+15)/\sqrt{3}\Rightarrow\sqrt{3}x=x+15\Rightarrow x(\sqrt{3}-1)=15$.
4. So $x=\dfrac{15}{\sqrt{3}-1}=\dfrac{15(\sqrt{3}+1)}{2}$. Thus $H=\dfrac{15(\sqrt{3}+1)}{2}$, which is option **(b)**.

FINAL_ANSWER: <b>

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