Geometry

Quantitative Aptitude ( CCS)

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In the given figure, PQ is a diameter of the semicircle PABQ and O is its center. ∠AOB = 64°. BP cuts AQ at X. What is the value (in degrees) of ∠AXP ?

Explanation:

Given Diagram PQ is diameter of circle with center O, angle AOB=64

If we draw line AP Then
∠APB = 1/2∠AOB =32 (Angle at circumference with same arc is always half of the angle at center)
∠PAQ = 90 (angle at semicircle with diameter is always 90)
So,
∠AXP = 180 – (∠APB+∠PAQ)=180-(90+22) = 58°

By: Amit Kumar

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