Solution Arimuth 1=226^(@)Azimuth 2=221^(@)minimum length of spiral =40m.e=0.1quad(10%" superetevation) "V=70kmph". "Rodway width =9m(a).{:[e=0.004(V^(2))/(R)=>0.1=0.004 xx((70)^(2))/(R)],[=>R=(R=196(m))/(R)=(1145.916)/(196)=(1145.916 )/(R)=(5.84^(3) ... See the full answer