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. O 144.26 O 146.42 124.46 142.46
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. O 144.26 O 146.42 124.46 142.46
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. long
Compute the tangent distance of the curve.
144.26
O 146.42
124.46
O 142.46](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdc3d2f17-51e2-44b5-95fc-03e639cb4cc5%2Fc77dcdfa-7a9d-4d78-b331-943e6458d3c7%2F9bnl02f_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. long
Compute the tangent distance of the curve.
144.26
O 146.42
124.46
O 142.46
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