getting the inner corner angle

Question

I have four points that make concave quad:

now I wanna get the inner angle of the (b) corner in degrees. note: the inner angle is greater than 180 degree.

Answer 1

Join b-d splitting the angle into 2 parts.

Part 1: Using Cosine Rule: $$\cos(\theta_1)=\dfrac{ad^2-ab^2-bd^2}{2\times ab\times bd}$$

Similarly: $$\cos(\theta_1)=\dfrac{cd^2-bc^2-bd^2}{2\times bc\times bd}$$

Answer 2

Draw $ac$ and use the law of cosines at $\angle b$, then subtract from $360$

$226=68+50-2\sqrt{50\cdot 68} \cos \theta \\ \cos \theta\approx -0.926 \\ \theta \approx 157.83 \\ \text{Your angle } \approx 202.17$

Answer 3

Let $\alpha$ be the angle from $bc$ to $ba$ then $$\tan \alpha=\frac{m_2-m_1}{1+m_1m_2}$$ where $m_1$ is the inclination of $bc$ and $m_2$ is the inclination of $ba$.

$m_1=\frac{10-9}{18-11}=\frac{1}{7}$

$m_2=\frac{9-11}{11-3}=-\frac{1}{4}$

you want to find $360^o-\alpha$.

Edit: With a calculator:

$\alpha=202.17^o$