Question

In the given figure AB is the diameter of the semicircle APOB with centre O, < POQ = 48 ° cuts BP at X, calculate < AXP.

Answer 1

Given : AB is a diameter of the semicircle APQB with center O

POQ = 48°  AQ cuts BP at X

To Find : ∠AXP

Solution :

AB is Diameter

Hence  ∠APB = 90°   ( Angle subtended by diameter  at circle = 90°)

∠PAQ = (1/2)∠POQ   ( Angle subtended by Chord/arc at remaining circle  = (1/2) angle subtended at center of circle )

∠POQ   = 48°

=> ∠PAQ = (1/2)48°

=> ∠PAQ = 24°

in  Δ APX

∠APX = ∠APB = 90°   ( As X lies on PB)

∠PAX  = ∠PAQ = 24°   ( As X lies on AQ)

Sum of Angles of triangle = 180°

=> ∠APX + ∠PAX  + ∠AXP = 180°

=> 90° + 24° + ∠AXP = 180°

=> ∠AXP = 66°

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