# Question #d7dc2

##### 1 Answer
Jan 23, 2017

We know with reference to the given figure that by Law of parallelogram of vectors

$| \vec{R} | = \sqrt{| \vec{P} {|}^{2} + | \vec{Q} {|}^{2} + 2 | \vec{P} | \cdot | \vec{Q} | \cdot \cos \theta}$ .....(1)
and $\alpha = {\tan}^{-} 1 \left(\frac{| \vec{Q} | \sin \theta}{| \vec{P} | + | \vec{Q} | \cos \theta}\right)$ .....(2)

Inserting given values in equation (1) we get

$| \vec{R} | = \sqrt{| 5 {|}^{2} + | - 3 {|}^{2} + 2 | 5 | \cdot | - 3 | \cdot \cos 60}$
$\implies | \vec{R} | = \sqrt{23 + 9 + 2 \times 5 \times 3 \times 0.5}$
$\implies | \vec{R} | = \sqrt{47}$
$\implies | \vec{R} | = 6.86$, rounded to two decimal places