From point P on level ground, the angle of elevation to the... | SAT Right triangles and trigonometry Question, Answer Explained

Question bank / Right triangles and trigonometry

Math · Geometry and TrigonometryHard

From point $P$ on level ground, the angle of elevation to the top of a vertical tower is $45^\circ$ . From point $Q$ , which is $30$ meters closer to the tower along the same straight line, the angle of elevation to the top of the tower is $60^\circ$ . What is the height, in meters, of the tower?

$30$

$15 + 15\sqrt{3}$

$30\sqrt{3}$

$45 + 15\sqrt{3}$

Show answer and explanation

Answer: D

Let $h$ be the height and $d$ be the distance from $Q$ to the base of the tower. From $Q$ , $\tan 60^\circ = \frac{h}{d}$ , so $d = \frac{h}{\sqrt{3}}$ . From $P$ , $\tan 45^\circ = \frac{h}{d + 30} = 1$ , so $h = \frac{h}{\sqrt{3}} + 30$ . Multiplying by $\sqrt{3}$ gives $\sqrt{3}h - h = 30\sqrt{3}$ , so $h = \frac{30\sqrt{3}}{\sqrt{3} - 1} = \frac{30\sqrt{3}(\sqrt{3} + 1)}{2} = 45 + 15\sqrt{3}$ .

Choice C assumes $Q$ is $30$ meters from the base of the tower, but $30$ meters is the distance between $P$ and $Q$ .

Why the other choices are wrong

A. This would be the height if P were 30 meters from the tower, but 30 meters is the distance between P and Q.
B. This comes from an algebra slip that drops a factor of the square root of 3 when solving for h.
C. This assumes Q is 30 meters from the tower, but 30 meters is the distance between P and Q.

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