In the MNPQ diamond, the QNM angle is 80 degrees, find the QPM angle.

We are given the angle between the diagonal NQ and the side MN, <QNM = 80. Since the inner angles lying crosswise are equal for parallel lines MN and PQ, then <QNM = <NQP = 80.

But the triangle NPQ is isosceles, since the sides of the rhombus NP = QP are equal to each other. Then the angles at the base NQ are. <PQN = <PNQ = 80. Then the sought angle NPQ is 180 – 80 – 80 = 20 (degrees).

But it is required to find the angle QPM, which is equal to half the angle <NPQ, since the diagonal of the rhombus PN divides the angle of the rhombus in half. Then: <QPM = 1/2 (<NPQ) = 20/2 = 10 (degrees).

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