b) Two straight alignment of a road intersect at a chainage (300 + 15). If the length of long chord is (600 + X) m and the radius of simple circular curve introduced is (620 + X) m, find- (i) the tangent distance, (ii) the length of the curve, (iii) the chainages at tangent points (iv) the apex distance (v) the versed sine. Assume the length of the chain is 20 m.
b) Two straight alignment of a road intersect at a chainage (300 + 15). If the length of long chord is (600 + X) m and the radius of simple circular curve introduced is (620 + X) m, find- (i) the tangent distance, (ii) the length of the curve, (iii) the chainages at tangent points (iv) the apex distance (v) the versed sine. Assume the length of the chain is 20 m.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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last three digit of id = 088
![b) Two straight alignment of a road intersect at a chainage (300 + 15). If the length of long
chord is (600 + X) m and the radius of simple circular curve introduced is (620 + X) m,
find- (i) the tangent distance, (ii) the length of the curve, (iii) the chainages at tangent
points (iv) the apex distance (v) the versed sine. Assume the length of the chain is 20 m.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd132e3ce-3828-42c4-9a42-30827b91b377%2Ff65eba74-7fe5-42c0-b562-b7954dcf7ba0%2Fkkbpf82_processed.jpeg&w=3840&q=75)
Transcribed Image Text:b) Two straight alignment of a road intersect at a chainage (300 + 15). If the length of long
chord is (600 + X) m and the radius of simple circular curve introduced is (620 + X) m,
find- (i) the tangent distance, (ii) the length of the curve, (iii) the chainages at tangent
points (iv) the apex distance (v) the versed sine. Assume the length of the chain is 20 m.
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