Rankine's method table, 150° intersection, 3°

Angle of intersection (interior) = 150°, so deflection angle $\Delta = 180^\circ - 150^\circ = 30^\circ$. The degree of curve is taken for the normal chord (arc) of 20 m.

Elements

$$\begin{aligned} R &= \frac{180\,C}{\pi D} = \frac{180\times 20}{\pi\times 3} = 381.97\ \text{m}\\ T &= R\tan\frac{\Delta}{2} = 381.97\tan 15^\circ = 102.35\ \text{m}\\ L &= \frac{\pi R\Delta}{180} = 200.00\ \text{m} \end{aligned}$$ $$\begin{aligned} \text{Chainage of T}_1 &= 1585.00 - 102.35 = 1482.65\ \text{m}\\ \text{Chainage of T}_2 &= 1482.65 + 200.00 = 1682.65\ \text{m} \end{aligned}$$

Rankine's deflection angles

$\delta = 1718.87\,c/R$ minutes for a chord $c$. First sub-chord $c_1 = 17.35$ m (to the next 20 m peg), last sub-chord $c_n = 2.65$ m.

Check: final deflection $= \Delta/2 = 15^\circ00'00''$ (computed 15°00′00″).

Answer: $R = 381.97$ m, $T = 102.35$ m, $L = 200.00$ m, T₁ at 1482.65 m, T₂ at 1682.65 m; deflection angles as tabulated.