SOLUTION: Two ships leave a harbor at the same time, traveling on courses that have an angle of 110 degrees between them. If the first ship travels at 26 miles per hour and the second ship t

Algebra ->  Trigonometry-basics -> SOLUTION: Two ships leave a harbor at the same time, traveling on courses that have an angle of 110 degrees between them. If the first ship travels at 26 miles per hour and the second ship t      Log On


   

Question 827997: Two ships leave a harbor at the same time, traveling on courses that have an angle of 110 degrees between them. If the first ship travels at 26 miles per hour and the second ship travels at 34 miles per hour, how far apart are the two ships after 2.9 hours?
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
After 2.9 hours one ship has traveled 2.9 * 26 = 75.4 miles and the other ship has traveled 2.9 * 34 = 98.6 miles.

It may help you understand if you draw a diagram:
  • Draw a triangle with a 110 degree angle in it (just guess the angle, it doesn't have to be exact). Label the angle as "A".
  • The vertex of this angle is the harbor.
  • Label one side of the 110 degree angle as 75.4. This side is the route traveled by the first ship.
  • Label the other side of the 110 degree angle as 98.6. This is the route traveled by the other ship.
  • Label the third side of the triangle as "a". This is the distance between the ships after 2.9 hours.
Our task then is to find the length of the third side of the triangle. The triangle is not a right triangle so we cannot use the Pythagorean Theorem. But we can use the Law of Cosines. (We can't use the Law of Sines because we do not know either of the angles opposite the known sides.) The Law of Cosines, with the known sides and angle in it, says:
a%5E2=%2875.4%29%5E2%2B%2898.6%29%5E2-2%2875.4%29%2898.6%29cos%28110%29
Simplifying...
a%5E2=5685.16%2B9721.96-2%2875.4%29%2898.6%29cos%28110%29
a%5E2=5685.16%2B9721.96-14868.88%28-0.3420%29 (cos rounded to 4 places)
a%5E2=5685.16%2B9721.96%2B5085.15696
a%5E2=20492.27696
Square root of each side (ignoring the negative square root since the side of the triangle cannot be negative):
sqrt%28a%5E2%29=sqrt%2820492.27696%29
a=sqrt%2820492.27696%29
a+=+143.1512 (rounded to 4 places)
So the ships are approximately 143 miles apart after 2.9 miles.