SOLUTION: Hi can you please help me with this problem. A ship leaves a port and travels with a bearing of 55 degrees for 100 km. It then travels 500 km on a course of 120 degrees. How far is

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Question 1140522: Hi can you please help me with this problem. A ship leaves a port and travels with a bearing of 55 degrees for 100 km. It then travels 500 km on a course of 120 degrees. How far is the ship from the port? In order to return to the port, what is the bearing the ship must take?
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
in the diagram below, the following occurs.

GH is the north south line for point A.

PM is the east west line for point A.

EF is the north south line for point B.

NO is the east west line for point B.

IK is the north south line for point C.

QL is the east west line for point C.

the bearing form point A to point B is 55 degrees which is the angle GAB.

the bearing from point B to point C is 120 degrees which is the angle EBC.

angle FBC is equal to 60 degrees because it is supplementary to angle EBC which is 120 degrees.

angle ABD is 55 degrees because it is complementary to angle BAD which is 35 degrees because it is complementary to angle GAB which is 55 degrees.

angle ABC is equal to 115 degrees because it is the sum of angles ABD and FBC which makes it the sum of 55 and 60 degrees.

side AB of triangle ABC is equal to 100 km because it is the distance from point A to point B.

side BC of triangle ABC is equl to 500 km because it is the distance from point B to point C.

you can use the law of cosines to find the length of side AC of triangle ABC.

the law of cosines says that b^2 = a^2 + c^2 - 2 * a * c * cos(B).

the law of cosines assumes that side a is opposite angle A and side b is opposite angle B and side c is opposite angle C.

to fit this formula, we let:

angle (B) = angle ABC = 115 degrees.
side b = AC = what we want to find.
side c = AB = 100 kilometers.
side a = BC = 500 kilometers.

b^2 = a^2 + c^2 - 2 * a * c * cos(B) becomes:

b^2 = 500^2 + 100^2 - 2 * 500 * 100 * cos(115).

solve for b^2 to get b^2 = 302261.8262.

solve for b to get b = 549.7834357.

that's the direct distance from point C back to the port at point A.

we can now use the law of sines to find angle C.

law of sines says that b / sin(B) = c / sin(C).

we know:

b = 549.7834357
c = 100
angle B = 115 degrees.

b / sin(B) = c / sin(C) becomes:

549.7834357 / sin(115) = 100 / sin(C)

solve to get sin(C) = c * sin(B) / b = 100 * sin(115) 549.7834357 = .1648481435.

solve to get angle C = arcsin(.1648481435) = 9.488412932 degrees.

since triangle BFC is a right triangle and angle FBC is equal to 60 degrees, then angle BCF is equal to 30 degrees because angle BCF is complementary to angle FBC.

angle ACF is equal to 30 degrees minus 9.488412932 degrees = 20.51158707 degrees.

angle ICA (shown as angle 4 in the diagram) is equal to 270 plus 20.51158707 = 290.51158707 degrees.

that's the bearing from point C to point A.

your solutions are (if i did this correctly):

the direct distance from point C back to the port at point A is 549.7834357 kilometers.

the bearing from point C back to the port at point A is 290.51158707 degrees.