See this blog for more information about regular polygons. This is a particularly challenging high level Quant problem.
First of all, you must remember that the sum of the angles in any n-sided polygon equals (n – 2)*180°. Thus, the angles in a hexagon add up to 720°, and in a regular hexagon, each full angle is 720°/6 = 120°.
Now, look at triangle LMN. LM = MN, so it's an isosceles triangle. The angle at M is 120°, so the other two angles, at L & N, must each be 30°. Thus, ∠MNQ = 30°.
Of course, triangle KLM must be identical to triangle LMN, so ∠LMQ also equals 30°. Because the big angle at M, ∠LMN, is 120°, this means that
(∠QMN) = (∠LMN) – (∠LMQ) = 120° – 30° = 90°
Thus, triangle QMN has a 30° angle and a 90° angle, so the third angle, ∠MQN, must equal 60°. Answer = (D).
Watch the lessons below for more detailed explanations of the concepts tested in this question.
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