Problem 03 - Simple Curve

$(R^2 - 4R + 4) + 1225 = R^2$

$4R = 1229$

$R = 307.25 \, \text{ m}$
 

Degree of curve (arc basis):
$\dfrac{20}{D} = \dfrac{2\pi R}{360^\circ}$

$\dfrac{20}{D} = \dfrac{2\pi (307.25)}{360^\circ}$

$D = 3.7^\circ$           [ D ]   answer