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प्रश्न
In the following figure, O is the centre of the circle, ∠ PBA = 42°.
Calculate:
(i) ∠ APB
(ii) ∠PQB
(iii) ∠ AQB
उत्तर
(i) In circle C(O, r)
AB is the diameter.
So, ∠ APB = 90° ....(Angle in a semicircle)
(ii) Now in Δ APB,
∠ PAB = 180° - (∠ APB + ∠ ABP)
∠ PAB = 180° - ( 90° + 42°)
∠ PAB = 180° - 132°
∠ PAB = 48°
∠ PQB = ∠ PAB = 48° ....(Angle of the same segment)
Hence,
∠ PQB = 48°
(iii) AQBP is a cyclic quadrilateral.
∠ APB + ∠ AQB = 180°
⇒ 90° + ∠ AQB = 180°
⇒ ∠ AQB = 180° - 90° = 90°.
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