Mathematics
AB is a diameter of the circle with center O. The tangent at a point P meets AB produced in Q, ∠PAQ = 34°, find angles ∠PBA and ∠PQA.
Circles
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Answer
We know that,
Angle in a semi-circle is a right angle.
∴ ∠APB = 90°
In △ APB,
⇒ ∠APB + ∠PAB + ∠PBA = 180°
⇒ 90° + 34° + ∠PBA = 180°
⇒ ∠PBA = 180° - 90° - 34° = 56°.
We know that,
Angle subtended by an arc at the center of the circle is twice the angle subtended by it on the circumference of the circle.
∴ ∠POB = 2∠PAB = 2 × 34° = 68°.
In △ POQ,
⇒ ∠OPQ + ∠PQO + ∠POQ = 180°
⇒ 90° + 68° + ∠PQO = 180°
⇒ 158° + ∠PQO = 180°
⇒ ∠PQO = 180° - 158° = 22°.
From figure,
⇒ ∠PQA = ∠PQO = 22°.
Hence, ∠PBA = 56° and ∠PQA = 22°.
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