The eyes of a basket ball player are 6 feet above the floor. He is at the free-throw line, which is 15 feet from the center of the basket rim. The center of the rim is 10 feet above the floor
Before shooting the player decides to step back a few steps. His angle of elevation is now 12.5o. How far from the center of the rim is his eye?
I used cos12.5=15/x
The rim is only 4 feet above his eyes, so
4/x = sin 12.5°
we don't really care how far horizontally he is from the basket. All that stuff about 15 feet and stepping back is just noise. The final picture is just a 12.5° angle with a 4-ft height difference.
To determine how far the basketball player's eye is from the center of the rim, we can use trigonometry.
First, let's consider the height of the eye above the floor. We are given that the center of the rim is 10 feet above the floor and the player's eyes are 6 feet above the floor. Therefore, the vertical distance between the center of the rim and the player's eye is 10 - 6 = 4 feet.
Next, we need to determine the horizontal distance between the center of the rim and the player's eye. To do this, we can use the tangent function, as we have the angle of elevation and the opposite side length (4 feet). The tangent of the angle of elevation is equal to the opposite side divided by the adjacent side.
tan(angle of elevation) = opposite/adjacent
tan(12.5°) = 4/adjacent
To find the adjacent side (the horizontal distance between the center of the rim and the player's eye), we can rearrange the equation:
adjacent = 4 / tan(12.5°)
Calculating this value, we find:
adjacent = 4 / tan(12.5°) ≈ 17.79 feet
Therefore, the player's eye is approximately 17.79 feet away from the center of the rim.