My first guess is that it could be found as first derivative of some function, but I don't have idea what that function could be.
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Welcome to MSE, Please show what you have thought or tried to solve the problem so that others will help you. This is how it goes on MSE. :) – Hosein Rahnama Jun 17 '16 at 11:58
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The required angle is given by $$\theta=\arctan\frac y2-\arctan\frac y8$$
When we differentiate and set to zero, we solve $$\frac{2}{4+y^2}=\frac{8}{64+y^2}\Rightarrow y=4$$ Which leads to $$\theta=\arctan2-\arctan\frac 12=\arctan \frac 34$$
David Quinn
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we first note that $$AB=6,BC=\sqrt{(64+y^2)},AC=\sqrt{(4+y^2)}$$ then we use the theorem of cosines $$36=4+y^2+64+y^2-2\sqrt{(4+y^2)}\sqrt{(64+y^2)}\cos(\gamma)$$ and from here we get $$\cos(\gamma)=\frac{(16+y^2)}{\sqrt{(4+y^2)}\sqrt{(64+y^2)}}$$ you must differentiate this with respect to $$y$$
Dr. Sonnhard Graubner
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