Law of Cosines Word Problem

Feb 2015
14
0
taytay rizal
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
Jun 2008
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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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Aug 2008
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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
Jun 2008
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North Texas
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.
 
Feb 2015
14
0
taytay rizal
How did you get the 35 and 40 degree?Thanks
 

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Aug 2008
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... 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.
 
Feb 2015
14
0
taytay rizal
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 :)
 

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MHF Helper
Aug 2008
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Surely you can see that $\displaystyle \begin{align*} 125^{\circ} - 90^{\circ} = 35^{\circ} \end{align*}$ and $\displaystyle \begin{align*} 270^{\circ} - 230^{\circ} = 40^{\circ} \end{align*}$...
 
Feb 2015
14
0
taytay rizal
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 :)
 

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MHF Helper
Aug 2008
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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?