#### 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.

#### Solution

Join OC.

Therefore, PA and PA 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° )

`∠`B = 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` × 120° = 60°

Is there an error in this question or solution?

Solution 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. Concept: Tangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments.