Suppose we have a triangle $ABC$ whose vertices are defined by the following matrix: $$\begin{vmatrix} A_x&A_y&1\\ B_x&B_y&1\\ C_x&C_y&1\\ \end{vmatrix}= \begin{vmatrix} 3&3&1\\ 0&0&1\\ 5&0&1\\ \end{vmatrix}$$
Is it possible to find the angles of the triangle using only the information from the matrix? By that I mean that we do not sketch the triangle. Or am I wrong? And the use of the distance formula is just indispensible?
On
Using the dot product of two vectors, we have: $$\vec{A}\cdot\vec{B} = AB\cos \alpha$$ First let's define the vectors we'd like to use: $$\vec{u}=\vec{AB}=\left<B_x-A_x,B_y-A_y\right>=\left<-3,-3\right>\\\vec{v}=\vec{AC}=\left<C_x-A_x,C_y-A_y\right>=\left<2,-3\right>$$ Now we can use the dot product, which is also defined to be the sum of the products of the components of the two vectors: $$\vec{u}\cdot\vec{v}=\vec{u}_x\vec{v}_x+\vec{u}_y\vec{v}_y=(-3)(2)+(-3)(-3)=3\\\vec{u}\cdot\vec{v}=3=|u||v|\cos\alpha\to\cos\alpha=\frac{3}{\sqrt{(-3)^2+(-3)^2}\sqrt{(2)^2+(-3)^2}}=\frac{1}{\sqrt{26}}\\\therefore\alpha=\cos^{-1}\left(\frac{1}{\sqrt{26}}\right)\approx78.69^\circ$$
Using law of cosines: $$\cos \angle C=\frac{(A_x-C_x)^2+(A_y-C_y)^2+(B_x-C_x)^2+(B_y-C_y)^2-(A_x-B_x)^2-(A_y-B_y)^2}{2\sqrt{[(A_x-C_x)^2+(A_y-C_y)^2]\cdot [(B_x-C_x)^2+(B_y-C_y)^2]}}$$