Question

In the figure, $\angle OAB={30}^o\&\angle OCB={57}^o.\angle AOC=?\&\angle BOC=?$

• A. $\angle AOC={120}^o\&\angle BOC={66}^o.$
• B. $\angle AOC={54}^o\&\angle BOC={57}^o.$
• C. $\angle AOC={30}^o\&\angle BOC={57}^o.$
• D. $\angle AOC={54}^o\&\angle BOC={66}^o.$

Answer 1

The correct option is D $\angle AOC={54}^o\&\angle BOC={66}^o.$

In the figure given, $\Delta OAB$ is an isosceles triangle since two of its sides $OA,OB$ are equal, both being the radii of the circle.
Given $\angle OAB={30}^o,$ we get $\angle OBA={30}^o.$
Also, since the sum of all the angles in a triangle is ${180}^o,\angle AOB={120}^o$
Now, $\Delta OBC$ is also an isosceles triangle since $OB,OC$ are again the radii.
$\therefore \angle OCB=\angle OBC={57}^o$
Since the sum of al the angles in a triangle is ${180}^o,\angle BOC={180}^o-{57}^o-{57}^o={66}^o.$
$\angle AOB=\angle AOC+\angle BOC$
$\Rightarrow \angle AOC={120}^o-{66}^o={54}^o$