Question 532·Hard·Math

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In right triangle $PQR$ , $\angle Q = 90^\circ$ . The altitude from $Q$ to the hypotenuse $PR$ meets $PR$ at point $S$ .

The lengths of the altitude and one of the hypotenuse segments are

QS = 12, \quad PS = 9.

What is the value of $\tan \angle RQS$ ?

An altitude from the right angle to the hypotenuse creates smaller right triangles with matching acute angles. Use right-triangle similarity to move the requested angle to a triangle whose two legs are already known, then write tangent as opposite over adjacent. You don't need to find every missing side. Reversing the legs gives the tangent of a different angle.

Hints

Hint 1
An altitude meets the side it drops to at a right angle. That makes $\triangle PQS$ a right triangle. Compare its acute angles with the two parts of the right angle at $Q$ .
Hint 2
Two angles are complementary if they add to $90^\circ$ . Which angle in $\triangle PQS$ and which part of the angle at $Q$ both complete $\angle PQS$ to $90^\circ$ ? Matching those angles avoids finding another side.

Step-by-step

Approach 1: Match the angle to a triangle with known legs
Step 1
Find the matching angle
No Desmos needed. An angle match lets you use the two given legs. An altitude meets the opposite side at a right angle, so $\angle QPS+\angle PQS=90^\circ$ . Also, $QS$ splits the right angle at $Q$ , so $\angle RQS+\angle PQS=90^\circ$ . Angles that add to $90^\circ$ with the same angle are equal, so $\angle RQS=\angle QPS$ .
Step 2
Write tangent using the given legs
In right triangle $PQS$ , tangent means opposite leg divided by adjacent leg. From $\angle QPS$ , the opposite leg is $QS=12$ and the adjacent leg is $PS=9$ . Equal angles have equal tangents, so $\tan\angle RQS=\tan\angle QPS=\frac{QS}{PS}=\frac{12}{9}$ .
Step 3
Reduce the ratio
Divide the numerator and denominator by $3$ : $\frac{12}{9}=\frac{4}{3}$ . So $\tan\angle RQS=\frac{4}{3}$ . Reversing the legs gives the tangent of the other acute angle, not the one asked for. Choice B.

Approach 2: Find the missing hypotenuse piece
Step 1
Connect the altitude to the two pieces
The altitude makes two similar right triangles: their right angles match, and so do their acute angles. For this setup, the altitude squared equals the product of the two hypotenuse pieces. Use $PS=9$ and call the other piece $RS$ : $12^2=9(RS)$ .
Step 2
Solve for the missing piece
Type $12^2\sim9x_1$ in Desmos, with $x_1$ standing for $RS$ . The $\sim$ tells Desmos to find the unknown value; under PARAMETERS, it reports $x_1=16$ . So $RS=16$ .
Step 3
Choose the sides for the requested angle
In right triangle $RQS$ , $RS$ is across from $\angle RQS$ and $QS$ touches it. So tangent is opposite over adjacent: $\tan\angle RQS=\frac{RS}{QS}=\frac{16}{12}$ .
Step 4
Evaluate the tangent
Add $x_1/12$ in Desmos to evaluate that ratio. It displays $1.3333\ldots$ ; use its fraction button to see $\frac{4}{3}$ . That is the tangent of $\angle RQS$ . Choice B.

Question 532·Hard·Math

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