Question

In the adjoining figure 'O' is the center of circle, $\angle$CAO = ${25}^{\circ }$ and $\angle$CBO = ${35}^{\circ }$. What is the value of $\angle$AOB?

• A. ${55}^{\circ }$
• B. ${110}^{\circ }$
• C. ${120}^{\circ }$
• D. Data insufficient

Answer 1

The correct option is C

${120}^{\circ }$

In $\Delta AOC$,
OA=OC --------(radii of the same circle)

$\therefore \Delta AOC$ is an isosceles triangle
$\to \angle OAC=\angle OCA={25}^{\circ }$----- (base angles of an isosceles triangle )

In $\Delta BOC$,
OB=OC --------(radii of the same circle)
$\therefore \Delta BOC$ is an isosceles triangle
$\to \angle OBC=\angle OCB={35}^{\circ }$ -----(base angles of an isosceles triangle )

$\angle ACB={25}^{\circ }+{35}^{\circ }={60}^{\circ }$
$\angle AOB=2\times \angle ACB$ ----(angle at the center is twice the angle at the circumference)

= 2$\times {60}^{\circ }$
=${120}^{\circ }$