In the given figure, O is the center of the circumcircle of triangle ABC. Tangents at A and B intersect at T. If $∠ A T B = 80^{\circ}$ and $∠ A O C = 130^{\circ}$ , Calculate $∠ C A B$ . [4 MARKS]

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Solution

Concept: 1 Mark
Application: 3 Marks

OA = OC (radius)

$\Rightarrow ∠ O C A = ∠ O A C$

In $Δ O C A, ∠ O C A + ∠ O A C + ∠ A O C = 180^{\circ}$ (By Angle Sum Property)

$2 ∠ O A C + 130^{\circ} = 180^{\circ}$

$2 ∠ O A C = 50^{\circ}$

$∠ O A C = 25^{\circ}$

We know that TA = TB

$\Rightarrow ∠ T A B = ∠ T B A$

In $Δ A T B, ∠ T A B + ∠ T B A + ∠ A T B = 180^{\circ}$ (By Angle Sum Property)

$2 ∠ T A B + 80^{\circ} = 180^{\circ}$

$2 ∠ T A B = 100^{\circ}$

$∠ T A B = 50^{\circ}$

$∠ O A B = ∠ O A T - ∠ T A B$

$∠ O A B = 90^{\circ} - 50^{\circ}$

$∠ O A B = 40^{\circ}$

$∠ C A B = ∠ O A C + ∠ O A B$

$∠ C A B = 25^{\circ} + 40^{\circ}$

$∠ C A B = 65^{\circ}$

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