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प्रश्न
In the given figure, O is the centre of the circumcircle ABC. Tangents at A and C intersect at P. Given angle AOB = 140° and angle APC = 80°; find the angle BAC.
उत्तर
Join OC.
Therefore, PA and PC are the tangents
∴ OA ⊥ PA and OC ⊥ PC
In quadrilateral APCO,
∠APC + ∠AOC = 180°
`=>` 80° + ∠AOC = 180°
`=>` ∠AOC = 100°
∠BOC = 360° – (∠AOB + ∠AOC)
∠BOC = 360° – (140° + 100°)
∠BOC = 360° – 240° = 120°
Now, arc BC subtends ∠BOC at the centre and ∠BAC at the remaining part of the circle.
∴ `∠BAC = 1/2 ∠BOC`
`∠BAC =1/2 xx 120^circ = 60^circ`
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