1. Trigonometry: bearing problems

I don't even know where to start with this problem...I know that i will end up drawing a triangle....but I could really use some assistance...I think that I did number 33 right, but i'm not quite sure

2. Hello, aikenfan!

(31) A ship leaves port at noon and has a bearing of S24°W.
The ship sails at 20 knots.
How many nautical miles south and west will the ship have sailed by 6 pm?
Code:
                P
*
/|
/ |
/  |
/24°|
/    |
120 /     |
/      |
/       |
/        |
* - - - - *
A         S

The ship leaves the port $P$ at 12 noon and sails at 20 knots.
. . In 6 hours, it has sailed 120 nautical miles to $A.$

We have: . $\cos24^o \:=\:\frac{PS}{120}\quad\Rightarrow\quad PS \:=\:120\!\cdot\!\cos24^o \:=\:109.6254549$

. . Therefore, the ship sailed about 109.6 nautical miles south.

We have: . $\sin24^o \:=\:\frac{AS}{120}\quad\Rightarrow\quad AS \:=\:120\!\cdot\!\sin24^o \:=\:48.80839717$

. . Therefore, the ship sailed about 48.8 nautical miles west.

3. Thank you very much! I just have one more question...for number 32....how do i do this if it gives a time (rather than a distance)?

4. Hello again, aikenfan!

For number 32....how do i do this if it gives a time (rather than a distance)?

Well, they did give you a distance . . . indirectly.

The plane flew for 1.5 hours at 600 mph . . . That's 900 miles.