Question 12 Marks
Can we have a rotational symmetry of order more than 1 whose angle of rotation is
(i) $45^{\circ}$ ?
(ii) $17^{\circ}$ ?
(i) $45^{\circ}$ ?
(ii) $17^{\circ}$ ?
Answer
View full question & answer→We know that an angle of rotation in one complete turn is $360^{\circ}$. So, if given angle of rotation divides $360^{\circ}$ completely, then for that rotational symmetry of order more than 1 that angle will be possible, otherwise not possible.
(i) Yes, we can have a rotational symmetry of order more than 1 whose angle of rotation is $45^{\circ}$. [$\because 360^{\circ}$ is divisible by $45^{\circ}$ completely]
(ii) No, we cannot have a rotational symmetry of order more than 1 whose angle of rotation is $17^{\circ}$. $\left[\because 360^{\circ}\right.$ is not completely divisible by $\left.17^{\circ}\right]$
(i) Yes, we can have a rotational symmetry of order more than 1 whose angle of rotation is $45^{\circ}$. [$\because 360^{\circ}$ is divisible by $45^{\circ}$ completely]
(ii) No, we cannot have a rotational symmetry of order more than 1 whose angle of rotation is $17^{\circ}$. $\left[\because 360^{\circ}\right.$ is not completely divisible by $\left.17^{\circ}\right]$















