Question

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.

Answer 1

 
Join OC.

Therefore, PA and PC are the tangents

∴ OA ⊥ PA and OC ⊥ PC

In quadrilateral APCO,

∠APC + ∠AOC = 180°

${\displaystyle \Rightarrow }$ 80° + ∠AOC = 180°

${\displaystyle \Rightarrow }$ ∠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.

 ∴ ${\displaystyle \angle BAC=\frac{1}{2}\angle BOC}$

 ${\displaystyle \angle BAC=\frac{1}{2}\times {120}^{\circ }={60}^{\circ }}$