Ans: (a) $60^\circ$ (b) $36$ (c) $3$
-
$\begin{array}{cl}
& \angle AOB \\
= & 135^\circ – 75^\circ \\
= & 60^\circ
\end{array}$ - Note that $OA = OB =12$. Therefore $\angle OAB = \angle OBA$. Hence, we have
$\begin{array}{cl}
& \angle OAB \\
= & \dfrac{1}{2} (180^\circ – \angle AOB) \\
= & \dfrac{1}{2} (180^\circ – 60^\circ) \\
= & 60^\circ
\end{array}$Therefore, $\angle OAB = \angle OBA = \angle AOB = 60^\circ$. Hence, $\Delta OAB$ is an equilateral triangle. Therefore, the perimeter of $\Delta AOB$
$\begin{array}{cl}
= & 3 \times 12 \\
= & 36 \text{ cm}
\end{array}$ - The number of folds of rotational symmetry of $\Delta AOB$ is $3$.
