Question

please answer my question​

Answer 1

Answer:

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Answer 2

∠BQR=∠QAB =70 degree

(Angles in alternate segments are equal)

Since radius divides tangent perpendicularly

∠CQR=90 degree

so ∠CQB =90-70 =20 degree---(1)

let the radius intersecting at chord AB be C.

since radius divides chord perpendicularly

∠OCA= 90 degree

By Angle sum property

∠AQC= 180- ∠OCA-∠QAB =180-90-70= 20 degree---(2)

From (1) and (2)

∠CQB +∠AQC=∠AQB

20+20= ∠AQB

.: ∠AQB=40 degree

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