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Difficulty: Hard

The figure presents line segments MQ and NR that intersect at point P, where point N is above and slightly to the right of point M and point Q is above and slightly to the left of point R. Line segments MN, QR, and horizontal line segment M, R are drawn forming triangles MNR and QMR. The measure of angle QPR is labeled 60 degrees and the angle measure of MQR is labeled 70 degrees.

In the figure above, $line segment M Q$ and $line segment N R$ intersect at point P, $N P equals Q P$ , and $M P equals P R$ . What is the measure, in degrees, of $angle Q M R$ ? (Disregard the degree symbol when gridding your answer.)

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Explanation

The correct answer is 30. It is given that the measure of $angle Q P R$ is $60 degrees$ . Angle MPR and $angle Q P R$ are collinear and therefore are supplementary angles. This means that the sum of the two angle measures is $180 degrees$ , and so the measure of $angle M P R$ is $120 degrees$ . The sum of the angles in a triangle is $180 degrees$ . Subtracting the measure of $angle M P R$ from $180 degrees$ yields the sum of the other angles in the triangle MPR. Since $180 minus 120, equals 60$ , the sum of the measures of $angle Q M R$ and $angle N R M$ is $60 degrees$ . It is given that $the length of side M P equals the length of side P R$ , so it follows that triangle MPR is isosceles. Therefore $angle Q M R$ and $angle N R M$ must be congruent. Since the sum of the measure of these two angles is $60 degrees$ , it follows that the measure of each angle is $30 degrees$ .

An alternate approach would be to use the exterior angle theorem, noting that the measure of $angle Q P R$ is equal to the sum of the measures of $angle Q M R$ and $angle N R M$ . Since both angles are equal, each of them has a measure of $30 degrees$ .