#### Answer

a. 56$^o$
b. 140s

#### Work Step by Step

a. Let the flow of the river be in the x direction and across the river be the y direction.
Let S denote Swimmer, R the River, and B the Bank. The pair “SB”, for example, represents the swimmer’s motion relative to the bank.
$$ \vec{v_{SB}} = \vec{v_{SR}} + \vec{v_{RB}} $$
$$(0, V) = (0.60 cos \theta, 0.60 sin \theta) m/s + (0.50, 0) m/s$$
Equate the x-components to solve for the angle of 146.44$^o$, or 56$^o$ west of north.
b. Find the crossing time from the y-direction.
First find the velocity in the y-direction.
$$(0, V) = (0.60 cos \theta, 0.60 sin \theta) m/s + (0.50, 0) m/s$$
$$V = (0.60 m/s)sin 146.44 ^{\circ} + 0 = 0.332 m/s$$
The time to cross is t = (45 m)/(0.332m/s) =135.68s $\approx$ 140s