Two cruise ships left the same port. After two hours, Ship A is 30 miles west and 18 miles north of the port and Ship B is 10 miles west and 27 miles south of the port. If there is a tug boat located one-fifth of the way from Ship A to Ship B, find the location of the tug boat relative to the port.

30.83 miles

Step-by-step explanation:

Distance of ship A from port is x

X²= 18²+30²

X²= 324+900

X²=1224

X=34.9857 miles

Bearing from port

Tan tita= 18/30

Tita = tan^-1 3/5

Tita= 31°

N 31° E

Distance of ship B from port is y

Y²= 27²+10²

Y²= 729+100

Y²= 829

Y=28.7924 miles

Bearing from port

Tan tita= 2.2

Tita= 69.7°

90-69.7= 20.3

N 20.3°W

Distance between the two ships is z

Z²=34.9857²+28.7924²-2(24.9857)(28.7924)cos(31+69.7°)

Z²= 2053+267.1

Z²= 2320.1

Z=48.18 miles

One fifth of 48.18

=9.63 miles

Distance of the boat is D

D/sin 60 = 34.9857/sin 100.7

D=30.83 miles

answer: 4 meters

step-by-step explanation: to you understand 2/3 better you can do 1 divided by 3 = . then multiply that by the 2 which is .66666666666666667 then you would do 6 x .66666666666666667 which is 4.      

hope this .

it would not be increasing by nothing that i can see probably by 25 percent that is all i can sat\y for now if i have any thing else to say i will edit this.

step-by-step explanation: