In the given figure, PQ is the 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:
Considering minor arc AB, ∠APB = ½ ∠AOB = 32°
Also, ∠PAQ = 90° (Since PQ is the diameter)
In ∆PAX,
∠PAQ + ∠AXP + ∠XPA = 180°
⇒ 90 + ∠AXP + 32 = 180
⇒ ∠AXP = 58°.
Hence, option (b).
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