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.

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 = $\dfrac{1}{2}$∠BOC = $\dfrac{1}{2} \times 120°$ = 60°.

Hence, ∠BAC = 60°.