Ambiguous Case Help

**Braedong** · December 3rd 2011, 01:54 PM

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?

**skeeter** · December 3rd 2011, 02:40 PM

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?

**Soroban** · December 3rd 2011, 07:53 PM

Hello, Braedong!

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 $024^o.$

**skeeter** · December 4th 2011, 06:21 AM

how about this sketch? amazing what can be done w/ MS Paint ...

**Braedong** · December 4th 2011, 08:52 AM

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

**skeeter** · December 4th 2011, 09:05 AM

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

I agree w/ those solutions. Understand that my "sketch" is not to scale.

**Braedong** · December 4th 2011, 09:10 AM

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?

**skeeter** · December 4th 2011, 09:30 AM

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 = $180 - \theta$

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