# Ambiguous Case Help

• Dec 3rd 2011, 01:54 PM
Braedong
Ambiguous Case Help
Ship Navigation Three ships, A, B, and C, are sailing along the same straight line on a course of 240°. Ship A is at the front, and ship C brings up the rear. From ship A. the bearing of a navigation buoy is 025°. The buoy is 2.5km from ship A and 2km from ships B and C. To the nearest degree, what is the bearing of the buoy from ship C? from ship B?
• Dec 3rd 2011, 02:40 PM
skeeter
Re: Ambiguous Case Help
Quote:

Originally Posted by Braedong
Ship Navigation Three ships, A, B, and C, are sailing along the same straight line on a course of 240°. Ship A is at the front, and ship C brings up the rear. From ship A. the bearing of a navigation buoy is 025°. The buoy is 2.5km from ship A and 2km from ships B and C. To the nearest degree, what is the bearing of the buoy from ship C? from ship B?

Have you made a sketch of the ships' and buoy's relative positions?
• Dec 3rd 2011, 07:53 PM
Soroban
Re: Ambiguous Case Help
Hello, Braedong!

Quote:

Three ships, A, B, and C, are sailing along the same straight line on a course of 240°.
Ship A is at the front, and ship C brings up the rear.
From ship A. the bearing of a navigation buoy is 025°.
The buoy is 2.5km from ship A, and 2km from ships B and C.
To the nearest degree, what is the bearing of the buoy from ship C? from ship B?

Code:

P          Q                  R
. . . *          *                  *    W
. . . :          :      X          :    * - - - - - - - - - - - * O
. . . :          :      o          :                60d    *
. . . :          :    **  *  2    :                *
. . . :          :  * *      *    :          *
. . . :          :*  *2          *  :      *
. . . :        *:  *              o
. . . :      *  : *          *    C
. . . : 25d *    :*    *
. . . :  * 5d  o
. . . : *  *    B
o
A

The ships are at $\displaystyle A.B,C.$ . The buoy is at $\displaystyle X.$ . $\displaystyle AX = 2.5,\;BX = CX = 2$

$\displaystyle \angle W\!O\!A = 60^o,\quad \angle P\!AO = \angle QBO = \angle RCO = 30^o$

$\displaystyle \angle P\!AX = 25^o \quad\Rightarrow\quad \angle X\!AC = 5^o$

In $\displaystyle \Delta AXC$, Law of Sines: .$\displaystyle \frac{\sin(\angle XCA)}{2.5} \,=\,\frac{\sin5^o}{2}$

. . $\displaystyle \sin(\angle XCA) \,=\,0.108944678 \quad\Rightarrow\quad \angle XCA \:\approx\:6^o$

$\displaystyle \angle RCA = 150^o \quad\Rightarrow\quad \angle RCX = 150^o - 6^o \,=\,144^o$

Therefore, the bearing of the buoy from ship $\displaystyle C$ is $\displaystyle 216^o.$

Since $\displaystyle \Delta BXC$ is isosceles, $\displaystyle \angle XBC = \angle XCA = 6^o$

Hence: $\displaystyle \angle QBX \:=\:30^o - 6^o \:=\:24^o$

Therefore, the bearing of the buoy from ship $\displaystyle B$ is $\displaystyle 024^o.$

• Dec 4th 2011, 06:21 AM
skeeter
Re: Ambiguous Case Help
• Dec 4th 2011, 08:52 AM
Braedong
Re: Ambiguous Case Help
I probably should have left the answers but Ship C is at 286° and ship B is at 14°. That seems to conflict with Soroban's answer though (Worried)
• Dec 4th 2011, 09:05 AM
skeeter
Re: Ambiguous Case Help
Quote:

Originally Posted by Braedong
I probably should have left the answers but Ship C is at 286° and ship B is at 14°. That seems to conflict with Soroban's answer though (Worried)

I agree w/ those solutions. Understand that my "sketch" is not to scale.
• Dec 4th 2011, 09:10 AM
Braedong
Re: Ambiguous Case Help
I've determined that B(uoy)AC is 30° and angle BCB(uoy) is 102.6° but from there I don't know how to determine the bearings of C in correlation with the buoy. How do you do that?
• Dec 4th 2011, 09:30 AM
skeeter
Re: Ambiguous Case Help
Quote:

Originally Posted by Braedong
I've determined that B(uoy)AC is 30° and angle BCB(uoy) is 102.6° but from there I don't know how to determine the bearings of C in correlation with the buoy. How do you do that?

no ... B(uoy)AC = 35 degrees

let $\displaystyle \theta$ = angle B(uoy)CA

$\displaystyle \frac{\sin{\theta}}{2.5} = \frac{\sin(35^\circ)}{2}$

then note that angle B(uoy)BA = $\displaystyle 180 - \theta$