# Bearing problem: Sine and cosine

• Mar 17th 2013, 06:01 PM
KerkLu
Bearing problem: Sine and cosine
Two ship leave a port at the same time. The first ship sails on a course of 35° at 15 knots, while the second ship sails on a course of 125° at 20 knots. Find after 2 hours (a)distance between the ship (b)the bearing from the first ship to the second (c) and the bearing from the second ship to the first.

• Mar 17th 2013, 06:16 PM
Paze
Re: Bearing problem: Sine and cosine
Well if the first ship leaves at a 35 degree angle relevant to the port and the 2nd ship leaves at a 125 degree angle relevant to the port, then the angle on the triangle which the ships form, becomes 90 degrees (125-35). The distance after 2 hours on the first ship is 30 knots and the 2nd ship has sailed 40 knots.

Since you have a right triangle, you can now use the Pythagorean theorem to show that the distance between the ships is $\displaystyle \sqrt{30^2+40^2}$

Now you can get the bearings or the other angles of the triangle by using (arc)trig functions.

$\displaystyle tan^{-1}\left(\frac{40}{30}\right)=53.13$. Now we can use: $\displaystyle 180-53.13-90=36.87$.
• Mar 17th 2013, 06:41 PM
KerkLu
Re: Bearing problem: Sine and cosine
That was quick! :) Thank you very much :) I understood your solution completely :) Never thought i would get a reply so fast :)