Question

Question 20
In figure, $\angle OAB={30}^{\circ }and\angle OCB={57}^{\circ }.Find\angle BOCand\angle AOC$.

Answer 1

Given, $\angle OAB={30}^{\circ }and\angle OCB={57}^{\circ }$
In $\Delta AOB$,
AO = OB [both are the radius of the same circle]
$\Rightarrow \angle OBA=\angle BAO={30}^{\circ }$ [angles opposite to equal sides are equal]
$\Rightarrow \angle AOB+\angle OBA+\angle BAO={180}^{\circ }$ [by angle sum property of a triangle]
$\therefore \angle AOB+{30}^{\circ }+{30}^{\circ }={180}^{\circ }$
$\therefore \angle AOB={180}^{\circ }-2\left({30}^{\circ }\right)$
$\angle AOB={180}^{\circ }-{60}^{\circ }={120}^{\circ }$...........(i)

Now, in $\Delta OCB$,
OC = OB [both are the radius of a circle]
$\angle OBC=\angle OCB={57}^{\circ }$ [angles opposite to equal sides are equal]
$\angle COB+\angle OCB+\angle CBO={180}^{\circ }$ [by angle sum property of triangle]
$\angle COB={180}^{\circ }-(\angle OCB+\angle OBC)$
$\angle COB={180}^{\circ }-({57}^{\circ }+{57}^{\circ })$
$\angle COB=180-{114}^{\circ }={66}^{\circ }$

From Eq. (i) $\angle AOB={120}^{\circ }$
$\Rightarrow \angle AOC+\angle COB={120}^{\circ }$
$\Rightarrow \angle AOC+{66}^{\circ }={120}^{\circ }$
$\therefore \angle AOC={120}^{\circ }-{66}^{\circ }={54}^{\circ }$