# Law of Cosines Word Problem

#### awts123

Can anyone help me to answer this one..Two ships leave the same port at the same time, One ships sails on a course of 125 degree at 18 knots while the other sails on a course of 230 degree at 24 knots.Find after 3 hours (a) the distance between a ship and (b) the bearing from the first ship to the second..Thanks

#### skeeter

MHF Helper
recall that bearings are measured clockwise from due North.

first ship on a bearing of 125 degrees 54 miles

second ship on a bearing of 230 degrees 72 miles

angle between the two bearings is 230-125 = 105 deg

part (a) is a rather straightforward application of the cosine law

for part (b), I'd find the measure of angle Q using the sine law and apply some basic geometry to determine the bearing

note the sketch is not to scale.

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#### Prove It

MHF Helper
recall that bearings are measured clockwise from due North.

first ship on a bearing of 125 degrees 54 miles

second ship on a bearing of 230 degrees 72 miles

angle between the two bearings is 230-125 = 105 deg

part (a) is a rather straightforward application of the cosine law

for part (b), I'd find the measure of angle Q using the sine law and apply some basic geometry to determine the bearing

note the sketch is not to scale.
Of course, these distances are nautical miles, which are different to ordinary miles...

#### skeeter

MHF Helper
Of course, these distances are nautical miles, which are different to ordinary miles...
... in over 20 years of service in the Navy, we just called them miles.

#### awts123

How did you get the 35 and 40 degree?Thanks

#### Prove It

MHF Helper
... in over 20 years of service in the Navy, we just called them miles.
Which is fine, but high school students don't have that experience and aren't normally able to make that distinction.

#### awts123

Which is fine, but high school students don't have that experience and aren't normally able to make that distinction.
can you help me to answer my problem ? Thanks #### Prove It

MHF Helper
Surely you can see that \displaystyle \begin{align*} 125^{\circ} - 90^{\circ} = 35^{\circ} \end{align*} and \displaystyle \begin{align*} 270^{\circ} - 230^{\circ} = 40^{\circ} \end{align*}...

#### awts123

Surely you can see that \displaystyle \begin{align*} 125^{\circ} - 90^{\circ} = 35^{\circ} \end{align*} and \displaystyle \begin{align*} 270^{\circ} - 230^{\circ} = 40^{\circ} \end{align*}...
Okay Thanks, but im still didnt know how to solve it.. i just know how to solve a triangle using law of cosine, but a word problem like this , its hard to me to understand.. can you give me more ideas to solve this.. Thanks #### Prove It

MHF Helper
Which part of the problem are you having trouble with? Skeeter already told you how to solve part 1. Can you see that in the triangle the angle between the two known lengths is \displaystyle \begin{align*} 105^{\circ} \end{align*}? Do you know how to apply the cosine rule \displaystyle \begin{align*} c^2 = a^2 + b^2 - 2\,a\,b\,\cos{(C)} \end{align*} to find an unknown length?