How is an exterior angle of a triangle related to the two interior angles that are not adjacent to it? Use that relationship to write an equation involving $5x + 20$, $2x - 10$, and $3(2x - 10)$.

Once you have an equation from the exterior angle theorem, solve for $x$. Then plug that value into the expressions for the interior angles of the triangle, and finally use either the triangle angle sum ($180^\circ$) or the fact that $\angle QRS$ and $\angle QRP$ form a linear pair to get $m\angle QRP$.

Be careful to notice that the problem asks for $m\angle QRP$, the interior angle at $R$, not the exterior angle $m\angle QRS$ and not the angle at $P$ or $Q$.

In one Desmos line, enter $y_1 = 5x + 20$. In another line, enter $y_2 = (2x - 10) + 3(2x - 10)$. These represent the two sides of the equation from the exterior angle theorem.

Use the graph to find the intersection point of $y_1$ and $y_2$. The x-coordinate of this intersection is the value of $x$ that satisfies the angle relationship.

In new lines, type expressions for the triangle’s angles using that x-value, for example: $2x - 10$ for $m\angle QPR$ and $3(2x - 10)$ for $m\angle PQR$. Desmos will show their numerical values.

Finally, in another line, type $180 - \big((2x - 10) + 3(2x - 10)\big)$ to use the triangle angle sum, or type $180 - (5x + 20)$ to use the linear pair at $R$. The numerical output of this expression is the measure of $\angle QRP$ in degrees.

Draw $\triangle PQR$ with side $PR$ extended past $R$ to a point $S$, forming exterior angle $\angle QRS$.

Label the given angles:

• $m\angle QPR = (2x - 10)^\circ$ (angle at $P$)
• $m\angle QRS = (5x + 20)^\circ$ (exterior angle at $R$)
• $m\angle PQR$ is three times $m\angle QPR$, so write it as $3(2x - 10)^\circ$ for now.

The exterior angle of a triangle equals the sum of the two remote (non-adjacent) interior angles.

Here, the exterior angle is $\angle QRS$, and the two remote interior angles are $\angle QPR$ and $\angle PQR$.

So:

Substitute the expressions for each angle:

First expand $3(2x - 10)$:

Now substitute back:

Combine like terms on the right side:

Solve the linear equation:

Subtract $5x$ from both sides:

Add $40$ to both sides:

Divide both sides by $3$:

Now substitute $x = 20$ into the expressions for the angles inside the triangle.

• $m\angle QPR = 2x - 10 = 2(20) - 10 = 40 - 10 = 30^\circ$
• $m\angle PQR = 3(2x - 10) = 3(30) = 90^\circ$

So the two known interior angles of $\triangle PQR$ are $30^\circ$ and $90^\circ$.

The sum of the interior angles of a triangle is $180^\circ$.

Let $m\angle QRP$ be the unknown angle at $R$. Then:

Combine the known angles:

Subtract $120$ from both sides:

So the measure of $\angle QRP$ is $60^\circ$. (You would enter $60$ as your answer.)