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Sagot :
Answer:
The part of the bridge that opens is 50 ft.
Step-by-step explanation:
The given parameters of the drawbridge are;
The entire length of the bridge = 100 feet
The height of the isosceles trapezoid formed = 25 feet
The angle at which the drawbridge meets the land ≈ 60°
Therefore, the part of the bridge that opens = The top narrow parallel side of the isosceles trapezoid
The length of each half of the bridge = (The entire length)/2 = 100 ft./2 = 50 ft.
Let 'x' represent the path of the waterway still partly blocked by each half of the bridge inclined
∴ x = 50 × cos(60°) = 25
x = 25 ft.
The path covered by both sides of the drawbridge = 2·x = 2 × 25 ft. = 50 ft.
The part of the bridge that opens = The entire length - 2·x
∴ The part of the bridge that opens = 100 ft. - 50 ft. = 50 ft.
The part of the bridge that opens = 50 ft.
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