# [SOLVED] Trig Bearing Problems

• Dec 30th 2008, 09:21 AM
gofish
[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...

• Dec 30th 2008, 11:10 AM
Soroban
Hello, gofish!

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

Quote:

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

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

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

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

• Dec 30th 2008, 01:14 PM
gofish
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 :D

I still need help on #2 though....
• Dec 30th 2008, 01:26 PM
skeeter
Quote:

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?
• Dec 30th 2008, 03:28 PM
gofish
Quote:

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
• Dec 30th 2008, 04:07 PM
skeeter
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.
• Dec 30th 2008, 05:34 PM
gofish
so that's how it's drawn! I didn't know where the wind was supposed be at.. thanks!! I understand it now.