SOLUTION: Two ships leave San Francisco at the same time. One travels N 40 W at a speed of 20 knots. The other travels S 10 W at a speed of 15 knots. How far apart are they after 11 hours? (

Algebra ->  Trigonometry-basics -> SOLUTION: Two ships leave San Francisco at the same time. One travels N 40 W at a speed of 20 knots. The other travels S 10 W at a speed of 15 knots. How far apart are they after 11 hours? (      Log On


   

Question 717432: Two ships leave San Francisco at the same time. One travels N 40 W at a speed of 20 knots. The other travels S 10 W at a speed of 15 knots. How far apart are they after 11 hours? ( 1 knot= 1 nautical mile per hour)
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
The distance from starting point of the mostly northward ship is 20*t, and the distance from starting point of the mostly southward ship is 15*t. If my understanding of the directions are correct, the angle of the two distance sides is 50%2B80 degrees. N 40 W may be 50 degrees northward from West and S 10 W may be 80 degrees southward from West. So I guess, the angle at starting point is 50%2B80=130 degrees.

A figure should be shown, but that is a long process here for me, yet I can feel a use for Law of Cosines. Using D for how far apart at 11 hours, the other side of the triangle, D%5E2=%2820t%29%5E2%2B%2815t%29%5E2-2%2820t%29%2815t%29cos%28130%29, and we want to finally let t=11.

D%5E2=400t%5E2%2B225t%5E2-600t%5E2%2Acos%28130%29
D%5E2=625t%5E2-600t%5E2%2Acos%28130%29
D%5E2=25t%5E2%2825-24%2Acos%28130%29%29
D%5E2=%2825t%5E2%29%2840.4269%29
D=5t%2Asqrt%2840.4269%29
highlight%28D=5%2At%2A%286.358%29%29, and LET t=11.