Answer
$v\approx\left\langle 38.751,27.982 \right\rangle$.
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 $|v|=47.8$ and $\theta=35^{\circ}50'=35^{\circ}+\left(\frac{50}{60}\right)^{\circ}=35^{\circ}+0.833^{\circ}=35.833^{\circ}$ since $1'=\left(\frac{1}{60}\right)^{\circ}$
therefore $\theta=35.833^{\circ}$.
Hence, the horizontal component is
\begin{align*}
a=&|v|\cos\theta\\
a=&47.8\cos35.833^{\circ}~~~ \text{use} ~~\theta=35.833^{\circ} ,|v|=47.8\\
a=&47.8\times 0.8107\\
a=&38.751 ~~~(\text{approximated to three decimal places}),
\end{align*}
and the vertical component is
\begin{align*}
b=&|v|\sin\theta\\
b=&47.8\sin35.833^{\circ}~~~ \text{use} ~~\theta=35.833^{\circ} ,|v|=47.8\\
b=&47.8\times 0.5854\\
b =&27.982~~~~~~ (\text{approximated to three decimal places}).
\end{align*}
Hence, the vector $v=\left\langle a,b \right\rangle\approx\left\langle 38.751,27.982 \right\rangle$ .
