The question states: Two towns A and B are located directly opposite each other on a river 8km wide which flows at a speed 4km/h. A person from town A wants to travel to a town C located 6km up-stream from and on the same side as B. The person travels in a boat with maximum speed 10km/h and wishes to reach C in the shortest possible time. Let x(t) be the distance travelled upstream and y(t) be the distance travelled across the river in t hours. The person heads out at angle theta. a) Show that x(t)=10tcos(theta)-4t and y(t)=10tsin(theta) b) What is the angle theta and how long would the trip take? Relevant equations: So far I have used v=d/t along with some vector diagrams. My attempt: I have proven a) already by using v=d/t. The net velocity for x was equal to 10cos(theta)-4 and I just rearranged for x. I did the same to find y. I then found the angle theta by saying that sin(theta)=8/10, therefore theta=arcsin(4/5). Also, I found the theta in terms of arccos which was theta=arccos(3/5). I found these by using a distance triangle with adjacent=6, opposite=8 and hypotenuse=10. I then equated x(t)=6 ==> 10tcos(theta)-4t=6 10tcos(arccos(3/5))-4t=6 10t(3/5)-4t=6 6t-4t=6 t=3 And equated y(t)=8 ==> 10tsin(theta)=8 10tsin(arcsin(4/5))=8 10t(4/5)=8 8t=8 t=1 This is where I'm having problems. Shouldn't the time value be equal? If anyone could please help me out I would greatly appreciate it.