Question

chords AB and CD of a circle intersect in point Q in the interior of a circle if m(arc AD) is equal to 25 degree and measure of Arc BC is equal to 30 degree then find angle BQC​

Answer 1

Answer:

Given−O is the centre of a circle.

Its chords AB&CD intersect at Q.

m(arcAD)=25° and m(arcBC)=31°.

To find out−∠BQC=?

Solution−

We join AD,BC&BD.

Also we joinAO,DO&BO,CO.

m(arcAD)=25°i.e∠AOD=25°.

Similarly m(arcBC)=31°i.e∠BOC=31°.

We know that the angle subtended by a chord of a circle at its centre is twice the angle subtended by the same chord at its curcumference.

∴∠BDC=21×∠BOC=21×31°=15.5°and∠ABD=21×∠BDC=21×25o=12.5o.

∴In ΔBQD we have ∠BQD=180°−(∠ABD+∠BDC)=180°−(12.5°+15.5°)=152°.

∠BQC=180°−152°=28°

Ans=28°