Question

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A boat leaves a buoy and travels at 12km/h on a bearing of 310T for 2 hours and then changes to travel on a bearing of 040T for 1 hour at 16km/h. What is the bearing of the buoy from the boat?

Answer 1

The bearing of an object is the angle of its position to a reference point. It can be determined by considering the clockwise sum of angles starting from the North pole. Thus, the bearing of the buoy to the boat is    $177^{o}$.

The speed of an object is defined as the rate of distance covered with respect to the time taken.

speed = $\frac{distance covered}{time taken}$

Thus, the first distance covered is given by:

Distance covered = speed x time taken

                              = 12 km/h x 2 h

                              = 24 km

The second distance covered is given as:

Distance covered = speed x time taken

                              = 16 km/h x 1 h

                              = 16 km

Applying the Cosine rule to determine distance from the boat to buoy, we have:

$c^{2}$ = $a^{2}$ + $b^{2}$ - 2abCos C

But, C = $50^{o}$ + $60^{o}$

          =      $110^{o}$

$c^{2}$  = $(16)^{2}$ + $(24)^{2}$ - 2(16)(24) Cos $110^{o}$

   = 256 + 576 - 768(-0.3420)

  = 832 + 262.656

$c^{2}$ = 1094.656

c = $\sqrt{1094.656}$

c  = 33.0856

c = 33.0 km

Apply the Sine rule to determine the angle at the stopping point of the boat;

$\frac{b}{Sin B}$ = $\frac{c}{Sin C}$

$\frac{24}{Sin B}$ = $\frac{33}{Sin 110}$

Sin B = $\frac{24*Sin 110}{33}$

         = 0.6834

B = 43.11

B = $43^{o}$

The angle made at the stopping point to the buoy is  $43^{o}$.

So that,

The bearing of the buoy from the boat = $180^{o}$- $3^{o}$

                                                                 =  $177^{o}$

Therefore, the required bearing of the buoy to the boat is   $177^{o}$.

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Answer 2

The bearing of an object is the angle of its position to a reference point. Thus, the bearing of the buoy to the boat is 117°.



What is speed?

The speed of an object is defined as the rate of distance covered to the time taken.

speed = Distance / time

Thus, the first distance covered is given by:

Distance covered = speed x time taken

                              = 12 km/h x 2 h

                              = 24 km

                             

The second distance covered is given as:

Distance covered = speed x time taken

                             = 16 km/h x 1 h

                             = 16 km

                           

Applying the Cosine rule to determine the distance from the boat to buoy, we have:

c² = a² + b² - 2abCos C

   = ( 16 )²  + ( 24 )² -2(16)(24)Cos110

   = 256 + 576 - 768(-0.3420)

   = 832 + 262.656

   = 1094.656

c² = 1094.656

c= √1094.656

c = 33.10 Km

Apply the Sine rule to determine the angle at the stopping point of the boat;

$\dfrac{b}{SinB}=\dfrac{c}{SinC}$

$\dfrac{24}{SinB}=\dfrac{33}{Sin110}$

Sin B = $\dfrac{24\times Sin110}{33}$

B = 43.11°

The angle made at the stopping point to the buoy is 43.11°  .

So that,The bearing of the buoy from the boat = 180 - 3 = 117°

Therefore the bearing of an object is the angle of its position to a reference point. Thus, the bearing of the buoy to the boat is 117°.



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