Question

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

Answer 1

Concept: 1 Mark
Application: 3 Marks

OA = OC (radius)

$\Rightarrow \angle OCA=\angle OAC$

In $\Delta OCA,\angle OCA+\angle OAC+\angle AOC={180}^{\circ }$ (By Angle Sum Property)

$2\angle OAC+{130}^{\circ }={180}^{\circ }$

$2\angle OAC={50}^{\circ }$

$\angle OAC={25}^{\circ }$

We know that TA = TB

$\Rightarrow \angle TAB=\angle TBA$

In $\Delta ATB,\angle TAB+\angle TBA+\angle ATB={180}^{\circ }$ (By Angle Sum Property)


$2\angle TAB+{80}^{\circ }={180}^{\circ }$

$2\angle TAB={100}^{\circ }$

$\angle TAB={50}^{\circ }$

$\angle OAB=\angle OAT-\angle TAB$

$\angle OAB={90}^{\circ }-{50}^{\circ }$

$\angle OAB={40}^{\circ }$

$\angle CAB=\angle OAC+\angle OAB$

$\angle CAB={25}^{\circ }+{40}^{\circ }$

$\angle CAB={65}^{\circ }$