Question

The resultant of two vectors of magnitudes 2A and $\sqrt{2}A$ acting at an angle $\theta$ is $\sqrt{10}A$. Find the value of $\theta$ ?

Answer 1

$R=\sqrt{a^2+b^2+2abcos\theta }$

$\sqrt{10}A=\sqrt{(2A{)}^2+\sqrt{2}{A}^2)+2\left(2A\right)\left(\sqrt{2}A\right)cos\theta }$

$10A^2=4A^{+}2A^2+4\sqrt{2}{A}^{2}cos\theta$

$10=6+4\sqrt{2}cos\theta$

$cos\theta =\left(\frac{4}{4\sqrt{2}}\right)=\left(\frac{1}{\sqrt{2}}\right)$

$\therefore \theta =\left(\frac{\pi }{4}\right)$