How would you find $x$ and $y$ coordinates of the third point in triangle($A$, $B$, $C$), if you know coordinates of $A$ and $B$, and angles at $A$ and $B$?
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1 Answer
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The slope described by an angle is given by $\tan(\theta)$. So let's suppose that $A = (a_1,a_2)$ and $B = (b_1,b_2)$
We get two lines:
$$y = \tan(\theta_A)(x-a_1) + a_2, y = \tan(\theta_B)(x-b_1) + b_2$$
$$\tan(\theta_A)(x-a_1) + a_2 = \tan(\theta_B)(x-b_1) + b_2$$
$$\tan(\theta_A)x-\tan(\theta_A)a_1 + a_2 = \tan(\theta_B)x-\tan(\theta_B)b_1 + b_2$$
$$\tan(\theta_A)x - \tan(\theta_B)x = -\tan(\theta_B)b_1 + \tan(\theta_A)a_1+ b_2 - a_2 $$
$$x(\tan(\theta_A) - \tan(\theta_B)) = -\tan(\theta_B)b_1 + \tan(\theta_A)a_1+ b_2 - a_2 $$
$$x = \frac{-\tan(\theta_B)b_1 + \tan(\theta_A)a_1+ b_2 - a_2}{(\tan(\theta_A) - \tan(\theta_B))} $$
Then substitute this into either equation:
$$y = \tan(\theta_A)(\frac{-\tan(\theta_B)b_1 + \tan(\theta_A)a_1+ b_2 - a_2}{(\tan(\theta_A) - \tan(\theta_B))} -a_1) + a_2$$
And so we have the ordered pair for the third point:
$$(\frac{-\tan(\theta_B)b_1 + \tan(\theta_A)a_1+ b_2 - a_2}{(\tan(\theta_A) - \tan(\theta_B))}, \tan(\theta_A)(\frac{-\tan(\theta_B)b_1 + \tan(\theta_A)a_1+ b_2 - a_2}{(\tan(\theta_A) - \tan(\theta_B))} -a_1) + a_2)$$