Math Help - Bearings....

1. Bearings....

a ship leaves port with a bearing of S 40[degrees] W. after traveling 7 miles, the ship turns 90 degress and travels on a bearing of N 50[degree] W for 11 miles. At that time, what is the bearing of the ship from port?

there's the picture that i drew for the diagram. sorry if it's confusing!

2. Originally Posted by CONFUSED_ONE
a ship leaves port with a bearing of S 40[degrees] W. after traveling 7 miles, the ship turns 90 degress and travels on a bearing of N 50[degree] W for 11 miles. At that time, what is the bearing of the ship from port?
Hello,

you have to calculate an angle in a right triangle. Let be $\alpha$ the angle at the point called port. Then you can use the tangens. And you'll get:
$\alpha=\arctan \left( \frac{11}{7} \right) \approx 57.5288^o \approx 57^o31'43.7''$

Now you have to caculate the bearing which is:
$180^o+40^o+57.5288^o=277.5288^o$

That means the bearing is now $W 7.5288^o N$. (A personal remark: When I worked as a seaman, we used numbers between $0^o (=N) \ to \ 360^o(=N)$ to name a bearing)

Greetings

EB

3. Hello earboth,
I thank you for solving this problem. In my calculations, the degree i got was 57 too. so that's good. what i don't understand is what you add to the degree to get the new bearing. why did you add 180 and 40 to it?

4. Originally Posted by CONFUSED_ONE
Hello earboth,
I thank you for solving this problem. In my calculations, the degree i got was 57 too. so that's good. what i don't understand is what you add to the degree to get the new bearing. why did you add 180 and 40 to it?
Hello,

as I've mentioned before I worked a s a seaman. The bearings we used are:

0° or 360° = N
90° = E
180° = S
270° = W

Substitute the letters in your bearing (S 40° W) you get 180°+40° because you go to W, that means your bearing is greater then 180°.

For me it's easier to calculate only with degrees instead of all the wests and easts and so on

Greetings

EB

5. Ohh..i see now. THANK you again.