Question: hello, please Help me to answer question 3(only), I am struggling a lot, you will see some formulas on the top. For the following horizontal
hello, please Help me to answer question 3(only), I am struggling a lot, you will see some formulas on the top.

For the following horizontal circular curve calculate the B.C. and E.C. chainages (km.stn.) and coordinates and calculate the deflection (d'), chord length and azimuth from the B.C. to ONLY the twelfth (12th) even station on curve. P.I.stn 7+500.000 km stn 2500.000 m North 2500.000 m East Back Tangent Azimuth 294 Ac 54 30' 45" (rotated counter-clockwise) R 500.000 metres Chaining interval 20.000 metres T = 257.586 m Lc = 475.711 m 180 T = R tan TRAC L = C: BCstn 7+242.414 km stn 2 180 2 T R 2395.230 m N 2735.317 m E C.L. = 2 R sind d = L are C Front Tangent Azimuth = 239.487500 ECstn 7+718.125 km stn 2369.217 m N 2278.085 m E C = 0.057295780 1st even station = 7+260.000 km stn 12th Even Station = 7+480.000 km stn Arc distance = 237.586 m Deflection (d) = 13.612675 Chord Length (C.L.) = 235.357 m Azimuth = 280.387325 N = 2437.665 m N E = 2503.817 m E
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