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?

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:

and for the other unknown:

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

dang that balls went far

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j