A ship leaves port at 10 miles per hour, with a heading of N 35° W. There is a warning buoy located 5 miles directly north of the port. What is the bearing of the warning buoy as seen from the ship after 7.5 hours?

Now I have drawn a diagram and solved that the angle should be 52.7 degrees I just can't figure out what direction. is it N 52.7 W
or would it be N 52.7 E

2. Re: Math problem me and friends are trying to solve

Hello, pierce!

I believe your diagram is wrong . . .

A ship leaves port at 10 mph with a heading of N 35° W.
There is a warning buoy located 5 miles directly north of the port.
What is the bearing of the warning buoy as seen from the ship after 7.5 hours?

Now I have drawn a diagram and solved that the angle should be 52.7 degrees.
. . Exactly where is that angle?
I just can't figure out what direction.
is it N 52.7 W or would it be N 52.7 E?
How can be the port be to the upper-left or upper-right?

After 10 hours, the ship has traveled 75 miles.
The diagram should look like this:

Code:
      X
*
:
:
:               Y
S *               *
: * *           :
:   *   *       :
:     *     *   :
*    75 *       * B
Z         *     |
*35d|5
* |
*
P

The port is at $P.$
The ship is at $S.$
The buoy is at $B.$
$\angle BPS = 35^o,\;SP = 75$

Law of Cosines:
. . $SB^2 \;=\;5^2 + 75^2 - 2(5)(75)\cos35^o \;=\;5035.635967$

. . . $SB \;=\;70.96221506 \;\approx\;71\text{ miles}$

Law of Sines:

. . $\frac{\sin(SBP)}{75} \:=\:\frac{\sin(SPB)}{71} \quad\Rightarrow\quad \sin(SBP) \:=\:\frac{75\sin(SPB)}{71} \:=\:0.605890602$

. . . . $\angle SBP \;=\;37.29295617 \;=\;37.3^o\text{ or }142.7^o$

We see that $\angle SBP$ is obtuse: . $\angle SBP \,=\,142.7^o$

Then: . $\angle XSB \,=\,142.7^o \quad\Rightarrow\quad \angle BSZ \,=\,37.3^o$

Therefore, the heading of the buoy from the ship is: . $S\,37.3^o\,E$