SAT Lines, Angles & Triangles Question 15 | Free SAT Prep

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Question 15·Hard·Lines, Angles, and Triangles

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Two straight conveyor belts are represented by parallel lines $\ell_1$ and $\ell_2$ . Point $P$ lies on $\ell_1$ , and points $Q$ and $R$ lie on $\ell_2$ . Segments $PQ$ and $PR$ are drawn.

The measure of the angle between $PQ$ and $\ell_1$ (on the interior side between the two parallel lines) is $(3x-20)^\circ$ . The measure of $\angle PQR$ is $(2x+10)^\circ$ , and the measure of $\angle PRQ$ is $(x+40)^\circ$ .

Which choice is the measure of $\angle QPR$ ?

When parallel lines appear, look for a transversal that creates equal (corresponding/alternate interior) angles so you can write an equation and solve for the variable. After that, treat the remaining work as a standard triangle problem: compute the needed angles and use the $180^\circ$ triangle angle sum to get the requested angle.

Hints

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Spot the parallel-line relationship

Because $\ell_1\parallel \ell_2$ , the angle that the transversal $PQ$ makes with $\ell_1$ matches the corresponding/alternate interior angle it makes with $\ell_2$ at $Q$ .

Set up an equation in $x$

Write an equation by setting $(3x-20)^\circ$ equal to $\angle PQR=(2x+10)^\circ$ , then solve for $x$ .

Finish with a triangle sum

After you know $x$ , compute $\angle PQR$ and $\angle PRQ$ , then use $\angle PQR+\angle PRQ+\angle QPR=180^\circ$ .

Desmos Guide

Solve for $x$ using the parallel-lines relationship

In Desmos, enter:

3x - 20 = 2x + 10

Then use Desmos to solve for $x$ .

Compute the two known angles in the triangle

Define:

Q = 2x + 10

R = x + 40

using the $x$ value you found.

Compute the requested angle

Define:

P = 180 - Q - R

The value of P is $\angle QPR$ . Only in this final step, match it to the answer choices.

Step-by-step Explanation

Use parallel lines to relate angles

Since $\ell_1\parallel \ell_2$ and $PQ$ is a transversal, the angle between $PQ$ and $\ell_1$ (on the interior side) equals the angle between $PQ$ and $\ell_2$ at $Q$ .

But the angle between $PQ$ and $\ell_2$ at $Q$ is $\angle PQR$ .

So,

(3x-20)^\circ=(2x+10)^\circ.

Solve for $x$

Solve the equation:

3x-20=2x+10 \implies x=30.

Find the two base angles of $\triangle PQR$

Substitute $x=30$ :

\angle PQR=(2x+10)^\circ=(2\cdot 30+10)^\circ=70^\circ

\angle PRQ=(x+40)^\circ=(30+40)^\circ=70^\circ.

Use the triangle angle sum to find $\angle QPR$

In $\triangle PQR$ , angles sum to $180^\circ$ :

\angle QPR=180^\circ-70^\circ-70^\circ=40^\circ.

So the correct choice is $40^\circ$ .

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation