Question

In fig, if $\angle ATO={40}^{\circ }$, find $\angle AOB$. [2 MARKS]

Answer 1

Concept: 1 Mark
Application: 1 Mark

As the lengths of tangents drawn from an external point to a circle are equal,

TA = TB

Also OA = OB (radius)

As the tangent at any point of a circle is perpendicular to the radius through the point of contact.

$\angle OBT=\angle OAT={90}^{\circ }$

Therefore $△OBT\cong △OAT$ by SAS Congruence Property

Hence $\angle BOT=\angle AOT\dots \left(i\right)$

Given that $\angle ATO={40}^{\circ }$

In right angled triangle AOT, By Angle Sum Property

$\angle AOT+\angle OTA+\angle OAT={180}^{\circ }$

$\angle AOT+{40}^{\circ }+{90}^{\circ }={180}^{\circ }$

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

From (i) $\angle BOT={50}^{\circ }$

$\angle AOB=\angle AOT+\angle BOT$

$\angle AOB={50}^{\circ }+{50}^{\circ }$

$\angle AOB={100}^{\circ }$