1. En dos estaciones de radio, A y C, que distan entre sí 50 km, son recibidas las señales que manda un barco, B. Si consideramos el

Question

1. En dos estaciones de radio, A y C, que distan entre sí 50 km, son
recibidas las señales que manda un barco, B. Si consideramos el
triángulo de vértices A, B y C, el ángulo en A es de 65º y el ángulo
en Ces de 809 ¿A qué distancia se encuentra el barco de cada una
de las dos estaciones de radio?

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Mít Mít 1 day 2021-07-22T06:51:36+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-07-22T06:53:34+00:00

    Answer:

    station A is 85.84 km from the ship, and station C is 79 km from the ship.

    Explanation:

    We can use the Law of Sines to solve this problem.

    Notice that in the triangle ABC (where B is the vertex representing the ship that is sending signals), the side opposite to angle B is denoted as “b”, and it is known to be 50 km (distance between vertex A and C).

    The side opposite to vertex A is denoted by “a” and is one of the unknowns = the distance between the boat and station C.

    Finally, the side opposite vertex C is denoted by “c” and is the other unknown = distance between the boat and station A.

    Given that we are dealing with a triangle, we know that the addition of the three internal angles should be 180 degrees, from which we can estimate the value of angle B: 180 – 65 – 80 = 35

    So now we use the law of sines to find sides a and c:

    \frac{50}{sin(35^o)} =\frac{c}{sin(80^o)} \\c=\frac{50\,*\,sin(80^o)}{sin(35^o)} \approx 85.84\,\,km

    and for the other unknown:

    \frac{50}{sin(35^o)} =\frac{a}{sin(65^o)} \\a=\frac{50\,*\,sin(65^o)}{sin(35^o)} \approx 79.00\,\,km

    Therefore, station A is 85.84 km from the ship, and station C is 79 km from the ship.

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