# Question #8f4d7

Sep 13, 2016

${55.8}^{\circ}$, rounded to one decimal place.

#### Explanation:

Given two vectors
$\vec{S}$, with $| \vec{S} | = 4$ and $\vec{R}$, with $| \vec{R} | = 6$
Let $\theta$ be the angle between the two.
By definition of dot product we have
$\vec{S} \cdot \vec{R} = | \vec{S} | | \vec{R} | \cos \theta$
Equating to the given value we get
$| \vec{S} | | \vec{R} | \cos \theta = 13.5$
$\implies 4 \times 6 \cos \theta = 13.5$
$\implies \cos \theta = \frac{13.5}{24} = 0.5625$
$\implies \theta = {\cos}^{-} 1 0.5625$
$\implies \theta = {55.8}^{\circ}$, rounded to one decimal place.