Explanation

(b) If $\left| \overrightarrow{A} \right| = \left| \overrightarrow{B} \right| = x,$ then $\left| \overrightarrow{C} \right| = \sqrt{2}x$
Now, $\overrightarrow{A} + \overrightarrow{B} = - \overrightarrow{C}$
or $\left( \overrightarrow{A} + \overrightarrow{B} \right) \bullet \left( \overrightarrow{A} + \overrightarrow{B} \right) = ( - \overrightarrow{C}) \bullet ( - \overrightarrow{C})$
or cos$\theta = 0$ or $\theta = 90{^\circ}$
or $\overrightarrow{A} \bullet \overrightarrow{A} + \overrightarrow{C} \bullet \overrightarrow{C} + 2\overrightarrow{A} \bullet \overrightarrow{C} = B^{2}$
or $x^{2} + 2x^{2} + 2x^{2}\sqrt{2}cos\theta = x^{2}$
or $cos\theta = - \frac{1}{\sqrt{2}}$
$\therefore$ or $\theta = 135{^\circ}$
Again, $\overrightarrow{B} + \overrightarrow{C} = - \overrightarrow{A}$
or ($\overrightarrow{B} + \overrightarrow{C}$)$\bullet \left( \overrightarrow{B} + \overrightarrow{C} \right) = - ( - \overrightarrow{A}) \bullet ( - \overrightarrow{A})$
or $x^{2} + 2x^{2} + 2x^{2}\sqrt{2}cos\theta = x^{2}$
or $cos\theta = - \frac{2x^{2}}{2x^{2}\sqrt{2}cos\theta} = - \frac{1}{2}$ or $\theta = 135{^\circ}$