Parametric equations motion problem

[cjh12398]

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 traveled upstream and y(t) be the distance traveled 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.

 

[the-ever-kid]

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.

the main reson that your time is not common is because the angle you found was in correct .

http://puu.sh/otUJ

solve for n and you get an angle that will give you time root 2 for both occasions