SOLUTION: Two ships leave a harbor entrance at the same time. The first ship is traveling at a constant 10 miles per hour, while the second is traveling at a constant 14 miles per hour. If t

Algebra ->  Triangles -> SOLUTION: Two ships leave a harbor entrance at the same time. The first ship is traveling at a constant 10 miles per hour, while the second is traveling at a constant 14 miles per hour. If t      Log On


   

Question 1136159: Two ships leave a harbor entrance at the same time. The first ship is traveling at a constant 10 miles per hour, while the second is traveling at a constant 14 miles per hour. If the angle between their courses is 123°, how far apart are they after 30 minutes? (Round your answer to the nearest whole number.)

= mi
Answer by VFBundy(438) About Me  (Show Source):
You can put this solution on YOUR website!
*  *
 *     *
  *         *       C
   *             *
 5  *                  *
     *                      *          
      *  123°                    *
       ******************************
                 7

After 30 minutes (0.5 hours), the ship that has a speed of 10 mph has traveled 5 miles and the ship that has a speed of 14 mph has traveled 7 miles. Given is that the two ships are 123° apart. So, this all forms a triangle. (See above.) We are looking for C.

Use the law of cosines to find C:
c² = a² + b² - 2ab(cos c)

Using our numbers:
c² = 5² + 7² - 2(5)(7)(cos 123°)

c² = 25 + 49 - 70(cos 123°)

c² = 74 - 70(cos 123°)

c² = 74 - 70(-0.54464)

c² = 74 + 38.1248

c² = 112.1248

c = 10.5889

The two ships are 10.5889 miles apart after 30 minutes.