The sum of the interior angles of a quadrilateral is 360°. Therefore, we have: $(x+10) + (3x-6) + (2x-8) + x = 360$ $7x - 4 = 360$ $7x = 364$ $x = 52$
$m\angle K = (x+10)^\circ = (52+10)^\circ = 62^\circ$ $m\angle J = (3x-6)^\circ = (3(52)-6)^\circ = (156-6)^\circ = 150^\circ$ $m\angle M = (2x-8)^\circ = (2(52)-8)^\circ = (104-8)^\circ = 96^\circ$ $m\angle L = x^\circ = 52^\circ$
$m\angle J = 150^\circ$ $m\angle K = 62^\circ$ $m\angle L = 52^\circ$ $m\angle M = 96^\circ$
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