Two ships leave port at the same time. One travels at $5$ km/h on a bearing of $46$ degrees. The other travels at $9$ km/h on a bearing of $127$ degrees. How far apart are the ships after $2$ hours?
[Math] Bearings and distances
algebra-precalculustrigonometry
Related Solutions
I think "the bearing of M from O is $38$ degrees" means if you draw a ray due East from O and put a point Q on it then you have to rotate the segment OQ $38$ degrees counterclockwise to get it to lie along the segment OM. And I think the question of the bearing from P to M is, through what angle must we rotate a segment extending due East from P to get it to lie along the segment from P to M.
If I have all that wrong, then ignore what follows.
Draw a diagram with all these points and lines. Angle POM is $152-38=114$ degrees. Triangle POM is isosceles, so angle PMO is $33$ degrees. Let the line due East from P cross OM at W. Angle PWO equals angle WOX (vertical angles), so it's $38$ degrees. So angle PWM is $142$ degrees. That makes angle MPW $5$ degrees, and that's the bearing of M from P.
An alternative approach is to use the Law of Cosines which states that $c^2 = a^2 + b^2 - 2ab\cos C$, where $a,b,c$ are the sides of any triangle and $C$ is the angle opposite side $c$. Using the information given in the problem, the triangle we sketch (below) has sides $a=7$, $b=9$ and $c$ to be found for the answer.
$ $
With a little geometry knowledge, we see that $m\angle C = 80^{\circ}$. So we get that $c^2 = 7^2+9^2-2*7*9*\cos(80^{\circ}) \approx 108.12 \Longrightarrow c \approx 10.398$
Best Answer
So, this will make a triangle, with one point being the port, and the other two points being the points where both ships are after 2 hours.
So, the angle between the two ships is $127-46=81$ degrees apart.
Also, one ship is $10$ km out $(5km/h*2h)$ and the other is is $18$ km out $(9km/h*2h)$
So, we have a triangle with an angle of $81$ degrees, and one side is $10$km, and the other is $18$km. How can we find the other side? Law of cosines. the law of cosines is:
$$c^2=a^2+b^2-2ab\cos C$$
We know that $a=10$ and $b=18$ (or we can switch those two, it doesn't really matter) and that C=$81$ degrees. Can you solve what the distance between them is from here?