A simple curve has a central angle of 40°. The stationing at the point of curvature is equal to 10+060. The offset distance from the PT to the tangent line passing thru the PC. is 80 m. long Compute the tangent distance of the curve. 144.26 O 146.42 124.46 O 142.46
A simple curve has a central angle of 40°. The stationing at the point of curvature is equal to 10+060. The offset distance from the PT to the tangent line passing thru the PC. is 80 m. long Compute the tangent distance of the curve. 144.26 O 146.42 124.46 O 142.46
Chapter2: Loads On Structures
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![A simple curve has a central angle of 40°. The stationing at the point of curvature is equal to 10+060. The offset distance from the PT to the
tangent line passing thru the PC. is 80 m. Iong
Compute the tangent distance of the curve.
144.26
146.42
124.46
O 142.46](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffda66c9f-6a6b-498a-87b6-198146cc7b72%2F03b08971-cbee-4761-9958-bf37921c193a%2F2r1dm2_processed.png&w=3840&q=75)
Transcribed Image Text:A simple curve has a central angle of 40°. The stationing at the point of curvature is equal to 10+060. The offset distance from the PT to the
tangent line passing thru the PC. is 80 m. Iong
Compute the tangent distance of the curve.
144.26
146.42
124.46
O 142.46
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