contestada

Two ships leave the same port at the same time. Two hours later ship A has traveled 12 miles and ship B has traveled 8 miles.




If they are currently 16 miles apart, what is the angle between their paths?

Respuesta :

Anzanzio

Answer:

the angle between their paths is 100.8°

Step-by-step explanation:

From the given information, you can construct a triangle, just like the one in the figure.

We will use the Cosine Rule which is:

c² = b² + a² - 2 b c cos(θ)

where

  • c = 16 miles
  • b = 8 miles
  • a = 12 miles

Therefore,

2 b c cos(θ) = b² + a² - c²

cos(θ) = (b² + a² - c²) / 2 b c

θ = cos⁻¹( (b² + a² - c²) / (2 b c) )

θ = cos⁻¹( (8² + 12² - 16²) / 2(8)(16) )

θ = 100.8°

Therefore, the angle between their paths is 100.8°

Ver imagen Anzanzio

Assuming both ships were travelling on straight paths, the angle between their paths is 104.48° approximately

What is law of cosine?

Let there is triangle ABC such that |AB| = a units, |AC| = b units, and |BC| = c units and the internal angle A is of θ degrees, then we have:

[tex]a^2 + b^2 -2abcos(\theta) = c^2[/tex]

(c is opposite side to angle A)

For the given situation, referring the diagram attached below, the point P is the starting point of ships and the point Q and R are the two hours later position of the ships.

The angle between the paths is the internal angle P made by PQ and PR.

Using the cosine law, we get:

[tex]a^2 + b^2 -2abcos(\theta) = c^2\\\\\theta = \cos^{-1}(\dfrac{a^2 + b^2 - c^2}{2ab})\\\\\theta = \cos^{-1}(\dfrac{144 + 64 - 256}{192}) = cos^{-1}(-0.25) \approx 104.48^\circ[/tex]

Thus, the angle between their paths is 104.48° approximately.

Learn more about law of cosines here:

https://brainly.com/question/17289163

Ver imagen astha8579

Otras preguntas