A ship leaves port at 1:00 P.M. and travels S35°E at the rate of 24 mi/hr.

Question

A ship leaves port at 1:00 P.M. and travels S35°E at the rate of 24 mi/hr. Another ship leaves the same port at 1:30 P.M. and travels S20°W at 22 mi/hr. Approximately how far apart are the ships at 3:00 P.M.? (Round your answer to the nearest mile.)

___________ mi

Answer 1

The first ship travels 35° east of due south for 2 hours at 24 mi/hr (distance = 48 mi)

The second ship travels 20° west of due south for 1.5 hours at 22 mi/hr (distance = 33 miles)

So, the angle between the paths of the two ships is 20° + 35° = 55°

Let x = distance between the ships at 3 PM.

By the Law of Cosines, x² = 48² + 33² - 2(48)(33)cos55°

x² = 1575.9098

x ≈ 40 miles

Answer 2

about 40 miles apart at 3pm

20+35=55

48=2(24), 33= 22(1.5)

it's a SAS triangle 33, 55 degree, 48

use law of cosines

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

where a=33, b= 48, C = 55

c^2 = 33^2 +48^2 -2(33)(48)Cos55

= 1089+2304 -66(48)Cos55

= 1575.90985

c = sqr1575.90985

c= about 39.69773104

= about 40