Super-Hard SAT Geometry Question 11 | Free SAT Prep

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Question 11·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

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A vertical antenna $\overline{TO}$ has its base at point $O$ on level ground and its top at point $T$ . Points $P$ and $Q$ are on the ground such that $\overline{PO}$ is perpendicular to $\overline{QO}$ and $PQ=55$ feet. The angle of elevation from $P$ to $T$ is $a^\circ$ , and the angle of elevation from $Q$ to $T$ is $b^\circ$ . If $\sin a^\circ=\frac{4}{5}$ and $\tan b^\circ=1$ , what is the height, in feet, of the antenna?

Your answer

For a problem involving angles of elevation from two ground points, first treat each line of sight as part of its own right triangle. Convert each trig ratio into a relationship between the vertical height and a horizontal ground distance. Then use the given relationship between the ground points, here a right angle and a hypotenuse length, to connect those distances with the Pythagorean theorem.

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Use the sine ratio at $P$

In right triangle $POT$ , the antenna height is opposite angle $a^\circ$ , and $PT$ is the hypotenuse. Interpret $\sin a^\circ=\frac{4}{5}$ as a side ratio.

Use the tangent ratio at $Q$

In right triangle $QOT$ , tangent is opposite over adjacent. Use $\tan b^\circ=1$ to compare $TO$ and $QO$ .

Connect the two ground distances

The segments $PO$ and $QO$ are perpendicular, and $PQ$ is their hypotenuse. Apply the Pythagorean theorem to triangle $POQ$ .

Desmos Guide

Set up a verification equation

Algebra is faster for this problem. For a Desmos verification, let $x$ represent the positive antenna height. The trig information gives $PO=\frac{3x}{4}$ and $QO=x$ .

Graph both sides of the Pythagorean equation

Enter $y=\left(\frac{3x}{4}\right)^2+x^2$ and $y=55^2$ . Their intersections represent values of $x$ that satisfy the Pythagorean theorem for triangle $POQ$ .

Select the positive intersection

Click the intersection with positive $x$ -coordinate. The positive coordinate is $44$ , so the antenna height is $44$ feet.

Step-by-step Explanation

Relate the height to $PO$

Let the antenna height $TO$ be $h$ . In right triangle $POT$ , $\sin a^\circ=\frac{TO}{PT}=\frac{4}{5}$ . Therefore, the side lengths have the ratio $3:4:5$ , so $PO=\frac{3h}{4}$ .

Relate the height to $QO$

In right triangle $QOT$ , $\tan b^\circ=\frac{TO}{QO}=1$ . Thus, $QO=h$ .

Use the ground triangle

Since $PO$ is perpendicular to $QO$ , triangle $POQ$ is a right triangle with hypotenuse $PQ=55$ . Substitute the two leg lengths: $\left(\frac{3h}{4}\right)^2+h^2=55^2$ . This simplifies to $\frac{25}{16}h^2=3025$ , so $h^2=1936$ .

Choose the positive height

A height must be positive, so $h=44$ . Therefore, the height of the antenna is $44$ feet.

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation