Question 5·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

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A vertical antenna has base $B$ . Points $Q$ , $R$ , and $B$ lie on the same straight, level line in that order, and $QR=12$ meters. The angles of elevation from $Q$ and $R$ to the top of the antenna are acute complementary angles. The distance from $Q$ to the top of the antenna is $4\sqrt{35}$ meters. Which choice gives the height of the antenna, in meters?

With two complementary angles of elevation, the tangent values multiply to $1$ . That connects the antenna’s height to both ground distances. Use the given slanted distance in the Pythagorean theorem, then solve the resulting equation in Desmos. The key trap is treating a distance along the ground as though it were the slanted side of a right triangle.

Hints

Hint 1
Because $R$ lies between $Q$ and $B$ , the nearer ground distance is shorter than $QB$ by $12$ meters. If you call $QB$ $d$ , how would you write $RB$ ?
Hint 2
Tangent compares the height opposite an elevation angle with the ground distance next to it. The tangents of complementary angles multiply to $1$ . What does that tell you about the height?
Hint 3
The distance from $Q$ to the top is the hypotenuse, the side across from the right angle. Use the Pythagorean theorem with that distance, then replace the height squared using your tangent equation.

Step-by-step

Connect the angles, then solve in Desmos
Step 1
Name the two ground distances
Let $d=QB$ and let $h$ be the antenna’s height. Because $R$ lies between $Q$ and $B$ , $RB$ is $12$ meters shorter than $QB$ : $RB=d-12,\qquad d>12$ .
Step 2
Write the tangent ratios
Call the elevation angles at $Q$ and $R$ $\theta_Q$ and $\theta_R$ . Tangent is the opposite leg divided by the adjacent leg. The height is opposite each angle, so: $\tan\theta_Q=\frac{h}{d},\qquad \tan\theta_R=\frac{h}{d-12}$ .
Step 3
Use the complementary angles
Complementary angles add to $90^\circ$ . Their tangents are reciprocals, so their product is $1$ : $\tan\theta_Q\tan\theta_R=1$ . Put in the ratios from the previous step: $\frac{h}{d}\cdot\frac{h}{d-12}=1$ . Multiply by $d(d-12)$ : $h^2=d(d-12)$ . The height squared equals the product of the two ground distances.
Step 4
Use the slanted distance
The vertical antenna meets the level ground at a right angle, so the distance from $Q$ to the top is the hypotenuse, across from that angle. Apply the Pythagorean theorem: $d^2+h^2=(4\sqrt{35})^2$ . Don’t use $12$ as a perpendicular leg; $QR$ lies along the ground.
Step 5
Make one equation in $d$
Replace $h^2$ with $d(d-12)$ from the complementary-angle step: $d^2+d(d-12)=(4\sqrt{35})^2$ . Now $d$ is the only unknown.
Step 6
Find the farther ground distance
Type $x_1^2+x_1(x_1-12)\sim(4\sqrt{35})^2\{x_1>12\}$ in Desmos. Here $x_1$ stands for $d$ , $\sim$ asks Desmos to solve, and the restriction keeps the distance longer than $12$ . Under PARAMETERS, Desmos shows $x_1=20$ , so $QB=20$ meters.
Step 7
Find the exact height
Use $h^2=d(d-12)$ and take the positive square root, since height is positive: $h=\sqrt{20(20-12)}$ . Add $\sqrt{x_1(x_1-12)}$ in Desmos; it shows about $12.649$ . To match the exact answer choices, simplify the radical: $h=\sqrt{160}$ $h=\sqrt{16\cdot10}$ $h=4\sqrt{10}$ . The antenna is $4\sqrt{10}$ meters tall. Choice A.

Question 5·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

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