Consider two vectors A and B , each having a magnitude of 10 unit....

Given $|\vec{A}| = 10 and$|\vec{B}| = 10$<br><br>Angle of$\vec{A}$with the X-axis is$30^\circ$and angle of$\vec{B}$with the X-axis is$60^\circ$.<br><br>The angle between the two vectors is$\theta = 60^\circ - 30^\circ = 30^\circ$.<br><br>The magnitude of the resultant$\vec{R}$is given by:<br><br>$R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$<br><br>$R = \sqrt{10^2 + 10^2 + 2(10)(10) \cos 30^\circ}$<br><br>$R = \sqrt{200(1 + \cos 30^\circ)}$<br><br>Using the identity$1 + \cos \theta = 2 \cos^2 \left(\dfrac{\theta}{2}\right):

$R =$\sqrt{200 \left(2 \cos^2 15^\circ\right)} = \sqrt{400 \cos^2 15^\circ} =$20 \cos 15^\circ$

Since the magnitudes of both vectors are equal, their resultant bisects the angle between them.

Angle of the resultant with the X-axis is $30^\circ +$\dfrac{30^\circ}{2} =$45^\circ$.