If S is the length of a subchord and R is the radius of simple curve, the angle of deflection between its tangent and sub-chord, in minutes, is equal to

ANS:E - 1718.9 S/R.

I don't think this answer (E) is correct.
It is only correct if the sub chord, S, was actually the sub arc length. Otherwise, the real answer should be sin^-1 (S/2R)*60. Actually the angular method of curve setting ,by Rankine's deflection method the deflection angle b/w chord and point of tangency is =1720C/R i, where c is the first sub chord and R is the radius. The angle of deflection in radian = S/2R.
The angle of deflection in degree = S/2R × 180/π
The angle of deflection in minutes = S/2R × 180/π × 60.
= 1718.18 ×S/R. For an angle of deflection in MINUTES then use = (S/2R)*(180/π)* 60.
= S/R*((180*60)/(2*(22/7)).
=S/R *1718.18,
=1718.18S/R Angle of deflection= (360/4π)* S/R.
This gives answer 28.64 S/R.
which is in degrees,, soo one degree is 60 min.
Thus, 28.64 S/R degrees in radians is answer E) 1718.9 S/R.