Mathematics
In the given figure, O is the center of the circumcircle ABC. Tangents A and C intersect at P. Given angle AOB = 140° and angle APC = 80°; find the angle BAC.
Circles
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Answer
Join OC
∴ PA and PC are the tangents
∴ OA ⊥ PA and OC ⊥ PC
In quadrilateral APCO,
⇒ ∠APC + ∠AOC = 180°
⇒ 80° + ∠AOC = 180°
⇒ ∠AOC = 180° - 80°
⇒ ∠AOC = 100°
From figure,
∠BOC = 360° - (∠AOB + ∠AOC)
= 360° - (140° + 100°)
= 360° - 240° = 120°.
We know that,
The angle at the centre of a circle is twice the angle at the circumference, subtended by the same arc.
Now arc BC subtends ∠BOC at the centre and ∠BAC at the remaining part of the circle.
∴ ∠BAC = ∠BOC = = 60°.
Hence, ∠BAC = 60°.
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