Answer
(a) $14i+6\sqrt 3j$
(b) $17.4$ mi/h. N $53.4^{\circ}$ E
Work Step by Step
We can identify these vectors: river flow $\vec u=8i$, motorboat $\vec v=12cos(90-30)^{\circ}i+12sin(90-30)^{\circ}j=6i+6\sqrt 3j$
(a) The true velocity of the motorboat is the sum of the above vectors: $\vec v_t=\vec u+\vec v=14i+6\sqrt 3j$
(b) The true speed of the motorboat is given by $|\vec v_t|=\sqrt {(14)^2+(6\sqrt 3)^2}\approx17.4$ mi/h. The angle of the vector with respect to the horizontal axis is given by $\theta=tan^{-1}(\frac{6\sqrt 3}{14})=36.6^{\circ}$, thus the direction of the vector is N $53.4^{\circ}$ E
