Question

In the given figure, arc AB and arc BC are equal in length. If $\angle AOB={48}^o$, find
(i) $\angle BOC$
(ii) $\angle OBC$
(iii) $\angle AOC$
(iv) $\angle OAC$

Answer 1

The arc of equal lengths subtends equal angles at the centre



$\angle AOB=\angle BOC={48}^o$

$\angle AOC=\angle AOB+\angle BOC={48}^o+{48}^o={96}^o$

So the triangle formed △ BOC is an isosceles triangle with OB = OC as they are radii of the same circle.

$\angle OBC=\angle OCB$ [opposite angles of equal sides of an isosceles triangle]

We know that the sum of all the angles of a triangle is $={180}^o$

$\angle BOC+\angle OBC+\angle OCB={180}^o$

$2\angle OBC+{48}^o={180}^o[\angle OBC=\angle OCB]$

$2\angle OBC={180}^o–{48}^o$

$2\angle OBC={132}^o$

$\angle OBC={66}^o$

Here $\angle OBC=\angle OCB={66}^o$

Also, $\Delta$ AOC is an isosceles triangle with OA = OC as they are radii of the same circle

$\Rightarrow \angle OAC=\angle OCA$ [opposite angles of equal sides of an isosceles triangle]

We know that the sum of all the angles of a triangle is ${180}^o$

$\angle COA+\angle OAC+\angle OCA={180}^o$

Substituting the values

$2\angle OAC+{96}^o={180}^o[\angle OAC=\angle OCA]$

$2\angle OAC={180}^o–{96}^o$

$2\angle OAC={84}^o$

$\angle OAC={42}^o$

Here $\angle OCA=\angle OAC={42}^o$