Sta. P.T. of a compound curve is at 15 + 480.14. Intersection angle of the first curve I sub 1 = 30 degrees, the intersection angle of the second curve I sub 2 = 36 degrees, D sub 1 = 4 degrees, and D sub 2 = 5 degrees. What is the stationing of P.C.C.? Use arc basis. a. 15+ 342.5 O b. 15 + 336.14 O C. 15 + 325.28 d. 15+ 319.63
Sta. P.T. of a compound curve is at 15 + 480.14. Intersection angle of the first curve I sub 1 = 30 degrees, the intersection angle of the second curve I sub 2 = 36 degrees, D sub 1 = 4 degrees, and D sub 2 = 5 degrees. What is the stationing of P.C.C.? Use arc basis. a. 15+ 342.5 O b. 15 + 336.14 O C. 15 + 325.28 d. 15+ 319.63
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
Chapter4: Quadrilaterals
Section4.3: The Rectangle, Square, And Rhombus
Problem 42E: a Argue that the midpoint of the hypotenuse of a right triangle is equidistant from the three...
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![Sta. P.T. of a compound curve is at 15 + 480.14. Intersection angle of the first curve I sub
1 = 30 degrees, the intersection angle of the second curve I sub 2 = 36 degrees, D sub 1
= 4 degrees, and D sub 2 = 5 degrees. What is the stationing of P.C.C.? Use arc basis.
a. 15+ 342.5
O b. 15 + 336.14
O
C.
15 + 325.28
d. 15+ 319.63](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F57cfd381-4f1d-4250-b76d-a00280cd0214%2Fbea4f7ee-9587-4d15-8dca-9f4f5f96d58a%2Fdplrjs8_processed.png&w=3840&q=75)
Transcribed Image Text:Sta. P.T. of a compound curve is at 15 + 480.14. Intersection angle of the first curve I sub
1 = 30 degrees, the intersection angle of the second curve I sub 2 = 36 degrees, D sub 1
= 4 degrees, and D sub 2 = 5 degrees. What is the stationing of P.C.C.? Use arc basis.
a. 15+ 342.5
O b. 15 + 336.14
O
C.
15 + 325.28
d. 15+ 319.63
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