A ship travels at 25 knots from its home port on a course of 106degrees for 3 hours and then changes to a course of 196 degrees for 2 hours. Find the distance and bearing of the port to the ship.

Question

A ship travels at 25 knots from its home port on a course of 106degrees for 3 hours and then changes to a course of 196 degrees for 2 hours. Find the distance and bearing of the port to the ship.

This ship is ____ nautical miles away on a beating of ____ degrees

Answer 1

1 knot = about 1.15 mph

25 knots = anout 25(1.15)= 28.75 mph

in 3 hours the boat went 86.25 miles

in 2 hours it went 57.5 miles

distance from port = sqr(57.5^2 + 86.25^2) = side opposite 90 degreed right triangle= hypotenuse=h

= about 103.66

use the law of sines to find the angle opposite 57.5

sinA=57.5/103.66

A= sin^-1(57.5/103.66)= about 33.69 degrees

it helps to draw a diagram and the right triangle 196-106 = 90

add 33.69 to 106 = about 139.69 degrees= bearing

or work the problem in nautical miles, 1 nautical mile = about 1.15 regular miles

in 3 hours, the boat went 75 nautical miles

in 2 hours it went 50 natuical miles

hypotenuse = sqr(50^2 +75^2)= sqr(2500+5625) = sqr8125 = about 90.14 nautical miles from port

sinA =50/90.14

A = sin^-1(50/90.14) = about 33.69

33.69+106= 139.69