Triangle, Circles, Angles

Level 2

B C BC is the diameter of a semi-circle. The sides of A B AB and A C AC of a triangle A B C ABC meet the semi-circle in P P and Q Q respectively. P Q PQ subtends 14 0 ∘ 140^\circ at the centre of the semi-circle. Then ∠ A \angle A is equal to:

(A) 1 0 ∘ 10^\circ .
(B) 2 0 ∘ 20^\circ .
(C) 3 0 ∘ 30^\circ .
(D) 4 0 ∘ 40^\circ .

20 40 30 10

Amish Garg by Amish Garg , 20, India

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2 solutions

Sahar Bano
Apr 7, 2020

Let the centre of the semi circle be O

Now, as angle QOC=POB and POB+QOC+POQ=180°

=>QOC=(180°-140°)/2 = 20° In ∆QOC angle OQC = OCQ (Because opposite side OQ and OC are equal)

Also OQC + OCQ + QOC = 180°

=>OQC = (180° - QOC)/2 = 80°

Now as angle OQC + OQA = 180°

=> OQA = 180° - 80° = 100°

Now in quadrilateral OQAP, OQA+QAP+APQ+POQ=360°

=>2(OQA) + 140° + QAP = 360°. (OQA = OPA and POQ=140°)

=>QAP= 360°-200°-140°=20°

As QAP = A

Hence the answer is 20°

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Drex Beckman
Jan 28, 2016

The line BC op course forms a 180 degree angle. 180-140/2=20. By definition, there are two isoceles triangles formed opposite the 140 angle. These have one angle of 20 degrees, so we divide 180-20 by two to get 80. We can find the final angle now: 180-(80+80)=20.

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