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The eyes of a basketball player are 5 feet above the floor. The player is at the free throw line, which is 15 feet from the point on the floor directly below the center of the basket. The rim of a standard basketball hoop is 10 feet of the floor. What is the angle of elevation from the player's eyes to the center of the rim?

Sagot :

oladayoademosu

Answer:

18.44°

Step-by-step explanation:

The line of sight of the player to the rim, the height of the eyes of the player above the floor, the distance of the player from the center of the rim and the height of the center of the rim above the ground form a trapezium. The height  of the center of the rim above the ground and  the height of the eyes of the player above the floor form parallel sides.

Since The height  of the center of the rim above the ground = 10 ft and the height of the eyes of the player above the floor = 5 ft, the height of the triangular part of the trapezium is 10 ft - 5 ft = 5 ft. The base of this triangle is parallel to the side of distance of the player from the center of the rim = 15 ft. So the base of this triangle is 15 ft. Since the angle of elevation Ф of the line of sight of the player to the rim is adjacent to this base = 15ft and opposite to the height of the triangle = 5ft,

By trigonometric ratios,

tanФ =opposite/adjacent = 5 ft/15 ft = 1/3

Ф = tan⁻¹(1/3)

Ф = 18.435°

Ф ≅ 18.44°

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