# Math Help - [SOLVED] Trig Bearing Problems

1. ## [SOLVED] Trig Bearing Problems

#1 Starting at point A, a ship sails 18.5 km on a bearing of 189°, then turns and sails 47.8 km on a bearing of 317°. Find the distance of the ship from point A.

I'd asked for help on yahoo answers, and the people there told me this

x = 18.5cos 189° + 47.8cos 317° = 16.686
y = 18.5sin 189° + 47.8sin 317° = -35.494
distance = (16.686)² + (-35.494)² = 39.220 km

it's the correct answer in the book, but I dont understand where the equations came from..

#2 What bearing and airspeed are required for a plane to fly due north in 2.5 hours if the wind is blowing from a direction of 328° at 11 mph?

I know that the plane is currently at 160 mph now, but then I dont know how the picture would even look like so I'm not sure where to begin...

2. Hello, gofish!

I think it's easier with the Law of Cosines.

1) Starting at point A, a ship sails 18.5 km on a bearing of 189°,
then turns and sails 47.8 km on a bearing of 317°.
Find the distance of the ship from point A.
Code:
                      N
C *         M     :
\  *     :     :
\     * :     :
\      : *   :
\     :     * A
47.8 \    :    /:
\43°:   / :
\  :9°/9°:
\ : /   :
\:/    :
*     :
B     S

The ship sails from $A$ to $B\!:\;\;AB = 18.5$
Hence, major angle $NAB \,= \,189^o\quad\Rightarrow\quad \angle SAB = \angle ABM = 9^o$

Then it sails from $B\text{ to }C\!:\;\;BC = 47.8$
Major angle $MBC = 317^o\quad\Rightarrow\quad \angle MBC = 43^o$
. . Hence: . $\angle ABC = 52^o$

Law of Cosines: . $AC^2 \;=\;47.8^2 + 18.5^2 - 2(47.8)(18.5)\cos52^o \;=\; 1538.231115$

Therefore: . $A \:=\:39.22028959 \;\approx\;39.220$ km.

3. ohhh I see I see. I drew the picture in a different, more weird way. that's probably why I got confused. Thanks a ton

I still need help on #2 though....

4. Originally Posted by gofish
#2 What bearing and airspeed are required for a plane to fly due north in 2.5 hours if the wind is blowing from a direction of 328° at 11 mph?

I know that the plane is currently at 160 mph now, but then I dont know how the picture would even look like so I'm not sure where to begin...
how far north in 2.5 hrs?

5. Originally Posted by skeeter
how far north in 2.5 hrs?
ahh I forgot to put the distance! sorry

400 miles due north in 2.5 hrs

6. the triangle is a rough sketch of the navigation triangle for this problem.

AB is the wind vector ... it has a length of 11 mph

CB is the ground vector ... it has a length of 160 mph

angle B = 148 degrees

using the law of cosines, the air vector, CA = 169.4 mph

now use the law of sines to calculate angle C.

7. so that's how it's drawn! I didn't know where the wind was supposed be at.. thanks!! I understand it now.