In the given figure, O is the centre of the circumcircle of ΔABC. Tangents at A and B intersect at T. If ∠ATB = 80° and ∠AOC = 130°, calculate ∠CAB.

From figure,

TA and TB are the tangents.

∴ OA ⊥ TA and OB ⊥ TB

In quadrilateral AOBT,

By angle sum property,

⇒ ∠ATB + ∠TAO + ∠TBO + ∠AOB = 360°

⇒ ∠ATB + ∠AOB + 90° + 90° = 360°

⇒ ∠ATB + ∠AOB = 180°

⇒ 80° + ∠AOB = 180°

⇒ ∠AOB = 180° - 80°

⇒ ∠AOB = 100°.

From figure,

∠BOC = 360° - (∠AOC + ∠AOB)

= 360° - (130° + 100°)

= 360° - 230° = 130°.

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 ∠COB at the centre and ∠CAB at the remaining part of the circle.

∴ ∠CAB = $\dfrac{1}{2}$ ∠COB

= $\dfrac{1}{2}$ × 130° = 65°.

Hence, ∠CAB = 65°.