Question

Two ships leave port at the same time one trip takes a bearing of 26° east of the north The other ship takes a bearing of 32° west of the north both ships are traveling at a speed of 30 mph how far apart to the nearest mile will they be at the end of three hours

Answer 1

Answer:

$\boxed{\text{ 87 mi}}$

Step-by-step explanation:

The diagram below shows the general situation.

One ship is heading NWbN at 30 mi/h. The other is heading NNE at the same speed.

They have been travelling for 3 h, so they have each covered 90 mi.

The angle between their tracks is

∠A = 32 + 26 = 58°

We can use the Law of Cosines to calculate a, the distance between the ships.

a² = b² + c² - 2bccosA

   = 90² + 90² - 2 × 90 × 90cos58

   = 8100 + 8100 – 16 200 × 0.5299

   = 16 200 - 8585

   = 7615

a = 87 mi

After 3 h, the distance between the ships will be $\boxed{\textbf{ 87 mi}}$