Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
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.
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.
