Answer
(a) $10^4 (4.8i+ 0.4j )$
(b) $4.8\times10^4lb$. S $85.2$ E.
Work Step by Step
Use the figure given in the Exercise and represent the forces with vectors: $\vec F_1=10^4\langle 2cos(90-50)^{\circ}, 2sin(90-50)^{\circ} \rangle\approx10^4\langle 1.532, 1.286 \rangle$, and $\vec F_2=10^4\langle 3.4cos(75-90)^{\circ}, 3.4sin(75-90)^{\circ} \rangle\approx10^4\langle 3.284, -0.880 \rangle$
(a) The resultant force is the sum of the above two forces $\vec F=\vec F_1+\vec F_2\approx10^4\langle 4.8, 0.4 \rangle=10^4 (4.8i+ 0.4j )$
(b) The magnitude of the resultant force can be found as $|\vec F|=10^4\sqrt {(4.8)^2+(0.4)^2}\approx4.8\times10^4lb$. The angle of the force with respect to the horizontal axis is given by $\theta=tan^{-1}(\frac{0.4}{4.8})\approx4.8^{\circ}$, thus the direction is S $85.2$ E.
