Two ships leave a harbor at the same time, traveling on courses that have an angle of 110 ° between them.

Question

If the first ship travels at 30 miles per hour and the second ship travels at 40 miles per hour, how far apart are the two ships after 3 hours?

Answer 1

30 mph X 3 hours = 90 miles and

40 mph X 3 hours = 120 miles

therefore you have a triangle where a = 90, b = 120, c = ? and the angle C = 110 deg.

The law of cosines says c^2 = a^2 + b^2 - 2abCos(C)

therefore c^2 = 90^2 + 120^2 - 2*90*120*Cos(110)

c^2 = 8100 + 14400 - 21600*(-.342)

c^2 = 29887

so c = 172.9 miles between them

Answer 2

Draw the triangle. One side is 120 miles long, another is 90 miles long. The angle (A) between these two lines is 110 degrees (110*2*pi/360 radians).

You can use the cosine rule to determine the length of the diagonal:

a^2=b^2+c^2- 2bccosA

a=sqrt(90^2+120^2-2*90*120*cos(110*2*p...

a=172.88 miles