Question 328·Medium·Math

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A triangular sign is supported by two braces of equal length. In the figure, $\triangle PQR$ is isosceles with $PQ = PR$ . Segment $\overline{PQ}$ is extended through $Q$ to point $S$ . If $\angle SQR = 146^\circ$ , which choice is the measure of $\angle QPR$ ?

An exterior angle appears when a triangle’s side continues past a corner. Matching side marks signal an isosceles triangle, whose angles across from the equal sides match. Find the inside angle next to the exterior angle, then use the equal sides and the exterior-angle theorem to reach the requested angle. The tempting smaller answer may be a base angle, not the angle where the equal sides meet.

Hints

Hint 1
Two touching angles on a straight line total $180^\circ$ . The angle inside the triangle at $Q$ touches the marked outside angle. How can you find that inside angle?
Hint 2
In an isosceles triangle, equal sides face equal angles. Trace across from $PQ$ and $PR$ to find the matching angles; the angle at $P$ is not one of them.
Hint 3
An exterior angle equals the two inside angles away from its corner added together. Which angle must join the angle at $R$ to make the marked outside angle?

Step-by-step

Use the exterior angle and equal sides
Step 1
Find the angle inside at Q
The extension through $Q$ puts $P$ , $Q$ , and $S$ on a straight line. So the marked exterior angle $\angle SQR$ and the inside angle $\angle PQR$ add to $180^\circ$ . Subtract the outside angle to find the inside one:
$m\angle PQR=180^\circ-146^\circ=34^\circ.$
Step 2
Match the angle at R
The given $PQ=PR$ makes $\triangle PQR$ isosceles. Side $PQ$ faces the angle at $R$ , and side $PR$ faces the angle at $Q$ . Equal sides face equal angles, so:
$m\angle PRQ=m\angle PQR=34^\circ.$
Step 3
Relate the outside angle to the angle at P
The exterior-angle theorem says an outside angle equals the two inside angles away from its corner added together. The triangle and the straight angle both total $180^\circ$ and share the inside angle at $Q$ , so their remaining angles must match. Let $x_1$ be the measure of $\angle QPR$ in degrees. Using the angle at $R$ :
$146=x_1+34.$
Step 4
Solve for the requested angle
Type $146\sim x_{1}+34$ . In Desmos, $\sim$ asks for the value of $x_1$ that makes the sides match. Under PARAMETERS, Desmos shows $x_1=112$ . Since $x_1$ is the requested angle at $P$ , $m\angle QPR=112^\circ$ . Choice C.

Question 328·Medium·Math

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