Answer
The elbow exerts a force of $161.6N$ on the forearm at an angle of $12.1^{\circ} $ below the horizontal.
Work Step by Step
Let the +x-direction be to the right and +y be upwards with the elbow at the origin.
Since it is given that the forearm is at an angle of 43$^{\circ}$ from the horizontal and that the bicep is perpendicular to it, we can see that the bicep is at an angle of 133$^{\circ}$ from the horizontal (aka -47$^{\circ}$).
Resolving the forces of the bicep which lies at 133$^{\circ}$ from horizontal
$\vec{B}$ = 232 = -158N$\widehat{i}$ + 169.6 $\widehat{j}$
Similarly we can see that the total weight downwards is the sum of the weight of the forearm and the load
$\vec{W}$ = -21.5$\widehat{i}$ + -114$\widehat{i}$ = -135.5N $\widehat{i}$
Given that the sum of elbow force and the biceps force must balance the weight of the arm and the weight it is carrying, so their vector sum must be 135.5 N, upward
$\vec{B}$ + $\vec{E}$ = 0$\widehat{i}$ + 135.5$\widehat{j}$
-158N$\widehat{i}$ + 169.6 $\widehat{j}$ + $\vec{E}$ = 0$\widehat{i}$ + 135.5$\widehat{j}$
$\vec{E}$ = -158$\widehat{i}$ - 34.1$\widehat{j}$
The magnitude of the force given by the elbow can be calculated by,
$E$ = $\sqrt {158^{2}+ 34^{2}}$ = 161.6N
and the direction can be calculated by
$\theta$ = $\arctan (\frac{34}{158})$ = 12.1$^{\circ}$
