In the given figure, PQ is diamerter of the semicircle PABQ and O is its center. ∠AOB = 66°, BP cuts AQ at X. What is the value (in degrees) of ∠AXP?
Answer & explanation
Correct answer: option 2
∠AOB = 66°
⇒ ∠BPA = \(\frac{66}{2}=33°\) (angle by same chord)
⇒ ∠PAQ = 90° (PQ is a diameter)
So, ∠
PXA = 90° - 33° = 57°