Question

In the given figure, AB and CD are two equal chords of a circle, with centre O. If P is the mid-point of chord AB, Q is the mid-point of chord CD and ∠POQ = 150°, find ∠APQ.

Answer 1

It is given in the question that point.

P is the mid-point of the chord AB and Point Q is the mid-point of the CD.
⇒ ∠APO = 90°      ...( as the straight line drawn from the center of a  circle to bisect a chord, which is not a diameter, is at the right angle to the chord. )

As chords, AB and CD are equal therefore they are equidistant from the center i.e; PO = OQ      ...( ∵ Equal chords of a circle are equidistant from the center)

Now, the ΔPOQ is an isosceles triangle with OP = OQ as its two equal sides of an isosceles triangle.

The sum of all the angles of a triangle is 180°.
⇒ ∠POQ + ∠OPQ + ∠PQO = 180°
⇒ ∠OPQ + ∠POQ + 150° = 180° ...( Given: ∠POQ = 150° )
⇒ 2∠OPQ = 180° - 150°  ...( As, ∠OPQ = ∠PQO ) 
⇒ 2∠OPQ = 30°
⇒ ∠OPQ = 15°

As ∠APO = 90°
⇒ ∠APQ + ∠OPQ = 90°
⇒ ∠APQ = 90° - 15°     ....( As, ∠OPQ = 15° )
⇒ ∠APQ = 75°.