Question

A ship from Port travels fot 60km on the bearing of 129°.It continues to travel and changes direction for another 50km on a bearing of 080°

Answer 1

Answer:

The ship is approximately 100 km to the Port.

Step-by-step explanation:

The sketch to the question gives a triangle with two sides and an included angle. Let the distance from the Port to the stopping point of the ship be represented by c. It can be determined by the application of cosine rule. Cosine rule states that:

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

a = 50 km

b = 60 km

C = 51 + 80 = $131^{o}$

So that;

$c^{2}$ = $50^{2}$ + $60^{2}$ -2(50 x 60) Cos $131^{o}$

   = 2500 + 3600 -6000 x -0.6561

   =  6100 + 3936.6

$c^{2}$ = 10036.6

c = $\sqrt{10036.6}$

  = 100.183

c = 100.18 km

Thus, the ship is approximately 100 km to the Port.