Question

In the given figure, the lengths of arcs $ABandBC$ are in the ratio $3:2$.
If $\angle AOB={96}^o$, find
$\angle BOC$
$\angle ABC$

Answer 1

$\measuredangle AOB={96}^0$

$\overline{AB}:\overline{BC}=3:2$

$\Rightarrow \measuredangle AOB:\measuredangle BOC=3:2$

$\Rightarrow \frac{{96}^0}{\measuredangle BOC}=\frac{3}{2}\Rightarrow \measuredangle BOC=\frac{2}{3}\times {96}^0={64}^0$

Now, ${\Delta }^{\prime \prime }OBC$ is isosceles [$\because$ $OB=OC=r$]

$\therefore$ $\measuredangle BOC+2\measuredangle OBC={180}^0$

$\Rightarrow \measuredangle OBC=\frac{180-64}{2}={58}^0$

also ${\Delta }^{\prime \prime }AOB$ is isosceles [$because$ $OA=OB=r$]

$\therefore$ $\measuredangle AOB+2\measuredangle OBA={180}^0$

$\Rightarrow \measuredangle OBA=\frac{180-96}{2}={42}^0$

$\therefore$ $\measuredangle ABC={58}^0+{42}^0={100}^0$