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?
A. 36
B. 32
C. 58
D. 54
Answer: Option C
Solution (By Examveda Team)
∠AOB = 64°
Then,
∠BPA = 32°
∠PAQ = 90°
[∴ PQ is a diameter]
So, ∠PXA = 90 - 32 = 58°
This Question Belongs to Arithmetic Ability >> Geometry
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