Answer
$v\approx\left\langle 13.66,7.110 \right\rangle$ .
Work Step by Step
The horizontal component of a vector $v=\left\langle a,b \right\rangle$ having magnitude $|v|$ and direction angle $\theta$ is given by $a=|v|\cos\theta$.
Similarly, the vertical component is given by $b=|v|\sin\theta$.
Now we have $|v|=15.4$ and $\theta=27^{\circ}30'=27^{\circ}+\left(\frac{30}{60}\right)^{\circ}=27^{\circ}+0.5^{\circ}=27.5^{\circ}$ since $1'=\left(\frac{1}{60}\right)^{\circ}$
therefore $\theta=27.5^{\circ}$.
Hence, the horizontal component is
\begin{align*}
a=&|v|\cos\theta\\
a=&15.4\cos27.5^{\circ}~~~ \text{use} ~~\theta=27.5^{\circ} ,|v|=15.4\\
a=&15.4\times 0.887\\
a=&13.66~~~(\text{approximated to two decimal places}),
\end{align*}
and the vertical component is
\begin{align*}
b=&|v|\sin\theta\\
b=&15.4\sin27.5^{\circ}~~~ \text{use} ~~\theta=27.5^{\circ} ,|v|=15.4\\
b=&15.4\times 0.4617\\
b =&7.110~~~~~~ (\text{approximated to three decimal places}).
\end{align*}
Hence, the vector $v=\left\langle a,b \right\rangle\approx\left\langle 13.66,7.110 \right\rangle$ .
