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

A vertical object on level ground creates a right triangle from each viewing point. Translate the trigonometric ratios into side lengths, then join the ground lengths using the right angle between them. A familiar side pattern can save you from typing equations into Desmos. Keep the ground segment separate from the slanted lines up to the top; they belong to different triangles.

Hints

Hint 1
An angle of elevation is measured up from the ground. From $P$ , the antenna's height is opposite angle $a$ , while $PT$ is the hypotenuse, the side across from the right angle. Which two sides does sine compare?
Hint 2
The Pythagorean triple $3$ - $4$ - $5$ works because $3^2+4^2=5^2$ . If the height takes $4$ parts and $PT$ takes $5$ , how many parts does the ground segment $PO$ take?
Hint 3
The paths $PO$ and $QO$ meet at a right angle, so $PQ$ is the hypotenuse of the ground triangle. Tangent compares the height with $QO$ ; what does a tangent of $1$ tell you about those lengths?

Step-by-step

Use two 3-4-5 triangles
Step 1
Read the sine ratio from P
No Desmos needed. Both right triangles use the $3$ - $4$ - $5$ side pattern. The vertical antenna and level ground make triangle $POT$ right at $O$ . From $P$ , the height $TO$ is opposite angle $a$ , and $PT$ is the hypotenuse, the side across from the right angle. Sine is opposite over hypotenuse, so:
$\sin a^\circ=\frac{TO}{PT}=\frac45.$
Step 2
Find the ground length from P
The ratio gives $TO$ $4$ equal parts and $PT$ $5$ . The Pythagorean triple $3$ - $4$ - $5$ has $3^2+4^2=5^2$ , so the remaining leg $PO$ has $3$ parts. Call one part $k$ feet:
$TO=4k,\qquad PT=5k,\qquad PO=3k.$
Step 3
Find the ground length from Q
From $Q$ , tangent is opposite over adjacent. The height $TO$ is opposite angle $b$ , and ground segment $QO$ is adjacent, so the given tangent says:
$\tan b^\circ=\frac{TO}{QO}=1.$
A ratio of $1$ means these two legs are equal, so:
$QO=TO=4k.$
Step 4
Use the right triangle on the ground
Because $PO$ is perpendicular to $QO$ , triangle $POQ$ has legs $3k$ and $4k$ . The same $3$ - $4$ - $5$ pattern makes its hypotenuse $PQ$ equal to $5k$ . Since $PQ=55$ :
$5k=55.$
Don't put $55$ on $PT$ : $PQ$ lies on the ground, not along the line to the antenna's top.
Step 5
Find the length of one part
Divide by $5$ :
$k=11.$
Each of the equal parts used to describe both triangles is $11$ feet.
Step 6
Find the antenna's height
The height is the $4$ -part side $TO$ , not the $3$ -part ground side. Use $TO=4k$ :
$TO=4(11)=44.$
The antenna is $44$ feet tall. Grid in 44.

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

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