Question 48·Hard·Lines, Angles, and Triangles

0 / 50

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

For right-triangle angle-chasing problems with a segment parallel to one side, first use the fact that the two acute angles in a right triangle sum to $90^\circ$ to find any missing angle. Then use properties of parallel lines—especially corresponding and alternate interior angles—to match the new angles in the smaller, created triangles back to the original triangle. Focus on which lines form each angle rather than trying to visualize everything at once; matching sides and parallels systematically is much faster and less error-prone on the SAT.

Hints

Click me!

Use the right angle at E

In triangle $DEF$ , you know $\angle E = 90^\circ$ and $\angle F = 42^\circ$ . What must $\angle D$ be so that all three angles add up to $180^\circ$ ?

Compare ∠D and ∠FPQ

Think about which lines form $\angle D$ (at point $D$ ) and which lines form $\angle FPQ$ (at point $P$ ). How does the fact that $\overline{PQ} \parallel \overline{DE}$ help you compare these two angles?

Identify equal angles from parallel lines

Since $P$ lies on $\overline{DF}$ and $\overline{PQ} \parallel \overline{DE}$ , $\angle FPQ$ is formed by one side along $\overline{DF}$ and one side parallel to $\overline{DE}$ . Which angle in triangle $DEF$ is formed in the same way?

Combine your results

Once you know which angle in triangle $DEF$ equals $\angle FPQ$ , use your calculation from the first hint to get the measure of $\angle FPQ$ .

Desmos Guide

Compute the remaining acute angle

Once you have reasoned that $m\angle D = 90^\circ - 42^\circ$ (because the two acute angles in a right triangle sum to $90^\circ$ ), type 90-42 into the Desmos calculator and note the result.

Relate this to ∠FPQ

Use your geometric reasoning (from the parallel lines) that $m\angle FPQ = m\angle D$ , and take the numerical value you saw in Desmos for 90-42 as the measure of $\angle FPQ$ .

Step-by-step Explanation

Find the missing acute angle in the right triangle

In right triangle $DEF$ , $\angle E = 90^\circ$ and $m\angle DFE = 42^\circ$ .

In any triangle, the three angles add up to $180^\circ$ . So the angles at $D$ and $F$ (the two acute angles) must add up to $90^\circ$ :

m\angle D + m\angle F = 90^\circ

We are given $m\angle F = 42^\circ$ , so

m\angle D = 90^\circ - 42^\circ.

Relate ∠FPQ to ∠D using the parallel line

Point $P$ is on $\overline{DF}$ and point $Q$ is on $\overline{EF}$ , with $\overline{PQ} \parallel \overline{DE}$ .

Look at the angle $\angle FPQ$ :

One side is $\overline{PF}$ , which lies along $\overline{DF}$ .
The other side is $\overline{PQ}$ , which is parallel to $\overline{DE}$ .

Now look at $\angle FDE$ (the angle at $D$ in triangle $DEF$ ):

One side is $\overline{DF}$ .
The other side is $\overline{DE}$ .

Because $\overline{PF}$ is the same line as $\overline{DF}$ and $\overline{PQ} \parallel \overline{DE}$ , the angle between them is the same. So

m\angle FPQ = m\angle FDE = m\angle D.

Substitute the value of ∠D to find ∠FPQ

From Step 1, we found

m\angle D = 90^\circ - 42^\circ.

From Step 2, we know $m\angle FPQ = m\angle D$ , so

m\angle FPQ = 90^\circ - 42^\circ = 48^\circ.

So the measure of $\angle FPQ$ is $\boxed{48}$ .

Question 48·Hard·Lines, Angles, and Triangles

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation