Two points A and B are on opposite sides of a building. A surveyor selects a third point C to place a transit Point C is 47 feet from point A and 64 feet from point B. The angle ACB is 56 degrees. How far apart are points A and B?

Answer 1

We can use the law of cosines to solve for the distance between points A and B. The law of cosines states that for any triangle with sides of lengths a, b, and c and an angle C opposite the side of length c, the following equation holds:

c^2 = a^2 + b^2 - 2 * a * b * cos(C)

In this case, we have a = 47 ft, b = 64 ft, and angle C = 56 degrees.

We need to first convert the angle to radians:

56 degrees * (pi radians / 180 degrees) ≈ 0.977 radians

Now we can plug the values into the law of cosines formula:

c^2 = (47 ft)^2 + (64 ft)^2 - 2 * 47 ft * 64 ft * cos(0.977 radians)
c^2 ≈ 2209 + 4096 - (2 * 47 * 64 * 0.558)
c^2 ≈ 6305 - 3408.32
c^2 ≈ 2896.68

Now we can find the length of c (the distance between points A and B) by taking the square root:

c ≈ √2896.68
c ≈ 53.84 ft

So the distance between points A and B is approximately 53.84 feet.