# getting the inner corner angle

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.

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

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thank you very much, your idea as more useful because i don't wanna to ignore d point – MoHaKa Feb 22 '13 at 19:09
@MoHaKa: whatever you want, but paying attention to $d$ distracts from the problem at hand and doubles the work. – Ross Millikan Feb 22 '13 at 19:10
i realized that, but if i ignored (d) point I won't be able to determine where is the inner corner, isn't that? – MoHaKa Feb 22 '13 at 19:14
@MoHaKa: you gave us the coordinates. But if you mean looking at the shape of the quadrilateral to see which corner is concave, you are right. – Ross Millikan Feb 22 '13 at 22:14
yep, i meant that, anyway thank you for help. – MoHaKa Feb 23 '13 at 11:04

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

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

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Your approach is cleaner. – Inquest Feb 22 '13 at 18:58

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:

$\alpha=202.17^o$

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