Q44
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?
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
AnswerD.
54
Answer: Option C
Solution
Answer: Option C
Solution:
∠AOB = 64°
Then,
∠BPA = 32°
∠PAQ = 90°
[∴ PQ is a diameter]
So, ∠PXA = 90 - 32 = 58°