<s>[INST] Finding the largest triangle inscribed in the unit circle Among all triangles inscribed in the unit circle, how can the one with the largest area be found?
 [/INST] For the maximum area the normal height of the triangle must be maximum for a given base Hence, the triangle ABC inscribed in unit circle should be an isosceles triangle having, $\angle B=\angle C$,  base BC & angle between equal sides AB & AC be $\theta$ ($=\angle A$). Now, from right triangle, we get $$\sin\theta=\frac{\frac{BC}{2}}{1}\implies BC=2\sin\theta$$ Length of equal sides of triangle is determined as $$AB=AC=\frac{BC}{2\sin\frac{\theta}{2}}=\frac{2\sin\theta}{2\sin\frac{\theta}{2}}=2\cos\frac{\theta}{2}$$ Now, the area of isosceles $\Delta ABC$ is given as
$$A=\frac{1}{2}(AB)(AC)\sin \theta=\frac{1}{2}\left(2\cos\frac{\theta}{2}\right)^2\sin \theta=2\sin\theta\cos^2\frac{\theta}{2}=\sin\theta(1+\cos\theta)$$$$\implies A=\sin\theta+\frac{1}{2}\sin2\theta$$ Now differentiating the area (A) w.r.t. $\theta$ & equating it to zero, we get $$\frac{dA}{d\theta}=\cos\theta+\cos2\theta=0$$$$ \implies\cos\theta=-\cos2\theta=\cos(\pi-2\theta)$$ $$\implies \theta=\pi-2\theta \quad \text{or} \quad 3\theta=\pi \quad \text{or} \quad \theta=\frac{\pi}{3}=60^o=\angle A$$ $$\implies \angle B=\angle C=\frac{180^o-\angle A}{2}=60^o$$
Hence, the triangle having maximum area in a unit circle must be an equilateral triangle
($\color{#0ae}{\angle A=\angle B=\angle C=60^o}$) 
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