How Is the Second Leg of the Triangle Calculated in Vector Problem?

[NormalForce1]

Homework Statement

A student measures his own walking speed and discovers that it is 1.5 m/s. This student then plans a hike. He plans to start out by walking for 30 minutes due south at his usual constant speed. Then he will turn west slightly so that he is facing 35⁰ west of south, and he will walk (still at the same speed) for 45 minutes in that direction.

After he has completed this first 75 minutes of walking, If the student wants to walk back to where he started from, in what direction should he walk and how long will he need to walk for?

Relevant Equations

a^2 + b^2 = c^2
c^2 = sqrt{a^2+b^2-[2*a*b*cos(theta)]}

I tried finding the resultant vector which was -6360. The magnitude of -6360 is the distance the traveler must travel to reach the start.
I found the angle by using the triangular sum theorem on a right triangle that was split from a scalene triangle. The scalene triangle has side lengths of 6360, 2700, and 3660.

 

[PeroK]

Do you have a question?

 

[PhDeezNutz]

Where did you get 3660 from?

I agree with the 2700. Wouldn’t the second leg of the triangle equal 45 x 60 x 1.5?