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

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

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

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

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

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

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

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

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

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

• 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 $\theta$ = angle B(uoy)CA

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

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