Circular curve through D, 5 m from B

Step 1: Deflection angle

Direction of travel A→B: $90^\circ$. Bearing of CB is $290^\circ$, so the bearing of B→C is $290^\circ - 180^\circ = 110^\circ$.

$$\Delta = 110^\circ - 90^\circ = 20^\circ\ \text{(to the right)}$$

Step 2: Position of D

Back bearing BA $= 270^\circ$ and bearing BC $= 110^\circ$. The bisector of the interior angle ABC has bearing $\dfrac{270^\circ + 110^\circ}{2} = 190^\circ$, which is the bearing of BD. Hence D lies on the line from B to the centre, and BD is the external distance:

$$E = 5\ \text{m}$$

   A ------- B ------- (to C at 110°)
              \
               D  (BD = 5 m, 190°)
               |
               O

Step 3: Radius

$$E = R\left(\sec\frac{\Delta}{2} - 1\right)\ \Rightarrow\ R = \frac{5}{\sec 10^\circ - 1} = \frac{5}{1.01543-1} = 324.12\ \text{m}$$

Step 4: Tangent length and curve length

$$\begin{aligned} T &= R\tan\frac{\Delta}{2} = 324.12\tan 10^\circ = 57.15\ \text{m}\\ L &= \frac{\pi R\Delta}{180} = 113.14\ \text{m} \end{aligned}$$

Step 5: Setting-out angle for a 20 m chord

$$\delta = \frac{1718.87\,c}{R} = \frac{1718.87\times 20}{324.12} = 106.07' = 1°46′04″$$

Answer: $R = 324.12$ m, $T = 57.15$ m, $L = 113.14$ m, setting-out angle for a 20 m chord $= 1°46′04″$ (about 106.1′).