Trig Problem- Vectors 4 parts!?

This one got me stuck =P.


::A group of trig students want to cross a river that is 1500 feet wide and has a current of 4 mph flowing from north to south. Starting on the eastern shore, the group considers rowing their boat directly west at a still water rowing speed of 2 mph::





a. In what direction will they actually be traveling? (nearest tenth degree)





b. How far downstream from a point directly across the river will they land? (Round to nearest ft.)





c. If the group instead desires to go directly across and can increase their still water speed to 8 mph, then in what direction must they aim the boat. (Round to nearest tenth of a degree)





d. How long will it take them to go directly across the river given these conditions in c? *nearest tenth of a min.








*****!Many thanks in advance to anyone who hellps with this!~****

Trig Problem- Vectors 4 parts!?
a. Since the river is flowing at a rate of 4 mph and the boat is being rowed at 2 mph perpendicular to it they will be moving at an angle of tan(2/4) west of the direction the river is flowing in (don't know what direction you want to be 0 and whether angles increase clockwise or counter-clockwise).





tan(2/4) = tan(1/2) = 26.6 degrees west of south





b. this is a ratio:


(distance downstream)/(distance across) = (speed downstream)/(speed across)


D / 1500 = 4/2


D = 3000 feet





c. Set this up as a triangle. The longest side is 8 and it is pointing at an angle upstream to compensate for the speed of flow of the river. Since they want to move directly across, the base of the triangle is perpendicular to the river (east-west). So relative to this line, the angle is:


sin(A) = 4/8 = 1/2 = 30 degrees





So they must aim the boat at an angle of 30 degrees north of west.





d. The speed in the direction across the river is found from:


c^2 = a^2 + b^2


8^2 = 4^2 + b^2


64 = 16 + b^2


48 = b^2


b = 4SQRT(3) = 6.9 mph