<s>[INST] The sum of the lengths of the hypotenuse and another side of a right angled triangle is given. The sum of the lengths of the hypotenuse and another side of a right angled triangle is given.The area of the triangle will be maximum if the angle between them is:
$(A)\frac{\pi}{6}\hspace{1cm}(B)\frac{\pi}{4}\hspace{1cm}(C)\frac{\pi}{3}\hspace{1cm}(D)\frac{5\pi}{12}$
Let $a,b$ are sides of a right triangle other than hypotenuse.Then given that $\sqrt{a^2+b^2}+a$=constant=$k$
Area of triangle=$\frac{1}{2}ab=\frac{1}{2}a\sqrt{(k-a)^2-a^2}$
and then?
 [/INST] Let $\theta$ be the angle between the hypotenuse & one side of length say $x$ of a right angled triangle such that $0<\theta<\frac{\pi}{2}$ then we have $$\text{hypotenuse}=x\sec \theta$$
$$\text{another side}=x\tan \theta$$  
Now, according to the given condition, we have $$x\sec\theta+x=c\ \ \text{(any constant)}$$ $$x=\frac{c}{1+\sec\theta}$$ Now, the area say $A$ of right angled triangle is given as $$A=\frac{1}{2}(x)(x\tan \theta)=\frac{1}{2}x^2\tan \theta$$ Now, setting the value of $x$, we get $$A=\frac{1}{2}\left(\frac{c}{1+\sec\theta}\right)^2\tan \theta=\frac{c^2}{2}\frac{\tan \theta}{(1+\sec\theta)^2}$$ $$\frac{dA}{d\theta}=\frac{c^2}{2}\frac{(1+\sec\theta)^2(\sec^2\theta)-2\tan \theta(1+\sec\theta)(\sec\theta\tan \theta)}{(1+\sec\theta)^4}$$ For maximum value of $A$, we have $\frac{dA}{d\theta}=0$ hence $$\frac{c^2}{2}\frac{(1+\sec\theta)^2(\sec^2\theta)-2\tan \theta(1+\sec\theta)(\sec\theta\tan \theta)}{(1+\sec\theta)^4}=0$$ $$\sec\theta(1+\sec\theta)(\sec\theta+\sec^2\theta-2\tan^2\theta)=0$$ $$\sec\theta(1+\sec\theta)^2(2-\sec\theta)=0$$ $$\sec\theta \neq0$$ $$ 1+\sec\theta=0\iff \cos\theta=-1\iff \theta=\pi$$ but $0<\theta<\frac{\pi}{2}$ hence this solution is not acceptable. Now,   $$2-\sec\theta=0\iff \sec\theta =2\iff $$ $$\theta=\frac{\pi}{3}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{option (C)}:\ \ \frac{\pi}{3}}}$$
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