Question 392·Hard·Math

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In right triangle $DEF$ , $\angle E = 90^\circ$ . Point $P$ lies on $\overline{DF}$ , and point $Q$ lies on $\overline{EF}$ such that $\overline{PQ}$ is parallel to $\overline{DE}$ . If $m\angle DFE = 42^\circ$ , what is the measure, in degrees, of $\angle FPQ$ ?

Your answer

A segment parallel to a side of a right triangle makes a smaller right triangle. Trace the sides to find its right angle, then identify which angle it shares with the larger triangle. The other two angles are complementary, meaning they add to $90^\circ$ . Don't copy the given angle to the requested corner without checking where each angle sits.

Hints

Hint 1
The middle letter of an angle names its vertex, the point where its sides meet. Both $P$ and $Q$ lie on sides that meet at $F$ . Which angle in triangle $FPQ$ is the given angle at $F$ ?
Hint 2
Perpendicular lines meet at a right angle. Since $PQ$ is parallel to $DE$ and $QF$ runs along $EF$ , which corner of the smaller triangle has a right angle?

Step-by-step

Approach 1: Find the angles in the smaller triangle
Step 1
Identify the shared angle at F
No Desmos needed. The angle facts give the result directly. From $F$ , $P$ lies toward $D$ and $Q$ lies toward $E$ . So the angle at $F$ in triangle $FPQ$ is the given angle:
$m\angle PFQ=m\angle DFE=42^\circ.$
Step 2
Find the small triangle's right angle
Because $\angle E=90^\circ$ , $DE$ is perpendicular to $EF$ : they meet at a right angle. $PQ$ is parallel to $DE$ , and $QF$ runs along $EF$ , so the smaller triangle's right angle is at $Q$ :
$m\angle FQP=90^\circ.$
Step 3
Find the requested angle at P
The two non-right angles in triangle $FPQ$ are complementary: they add to $90^\circ$ . Subtract the angle at $F$ : $m\angle FPQ=90^\circ-42^\circ=48^\circ$ . So $\angle FPQ$ measures $48^\circ$ . Grid in 48.

Approach 2: Match angles across the parallel segments
Step 1
Find the angle at D
No Desmos needed. The acute angles of a right triangle add to $90^\circ$ . In triangle $DEF$ , $\angle E$ is right, so subtract the angle at $F$ :
$m\angle FDE=90^\circ-42^\circ=48^\circ.$
Step 2
Match D to P
$DF$ crosses the parallel segments $DE$ and $PQ$ . Angles $FDE$ and $FPQ$ are corresponding angles: each sits in the same position where $DF$ meets a parallel segment. So $P$ matches $D$ , not $F$ : $m\angle FPQ=m\angle FDE=48^\circ$ . The requested angle is $48^\circ$ . Grid in 48.

Question 392·Hard·Math

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