Largest possible angle formed by $(1,2)$, $(2,1)$ and vertex $(0,y)$ on the y-axis, given $y<3$ 
I solved it using law of cosine, then differentiate, but it seems really lengthy, does anyone have a better approach? please tell me.
$$\theta=cos^{-1}(\frac{a^2+b^2-c^2}{2ab})$$
 A: Note that if we have a circle with $A,B$ on it, a point $C$ on it views $AB$ at a constant angle.
we are looking for the smallest angle circle touching the y-axis, meaning it is tangent to it.
which is very clearly true when $C=(0,1)$ and thus the maximal angle is $45^o$
A: The angle $\theta$ can be calculated by subtracting the angle of elevation to $B$ from the angle of elevation to $A$.
Recall that the angle of elevation of a line passing with a slope $m=\dfrac{y_2-y_1}{x_2-x_1}$ is calculated by $\theta=\arctan m$.  Thus, $AC$ has an angle of elevation of $\arctan(2-y)$ and $BC$ has an angle of elevation of $\arctan\left(\frac{1-y}2\right)$.  Finding the difference allows us to use the subtraction angle formula for the arctangent function $$\arctan a-\arctan b=\arctan\left(\frac{a-b}{1+ab}\right).$$
Using the above, we have $$\theta=\arctan(2-y)-\arctan\left(\frac{1-y}2\right)=\arctan\left(\frac{3-y}{4-3y+y^2}\right).$$
To maximize $\theta$ as $y$ varies, we can find $y$ such that $\dfrac{d\theta}{dy}=0:$
$$\theta=\arctan\left(\frac{3-y}{4-3y+y^2}\right)\implies\frac{d\theta}{dy}=\frac{y^2-6y+5}{y^4-6y^3+18y^2-30y+25}$$ and it is trivial to see that $y=1$ satisfies zeroing the derivative.
