If the sum of the lengths of the hypotenuse and one of the other two sides of a right angled triangle is fixed, then the area of the triangle is maximum, when the angle between them is :

$\frac{\pi}{3}$

hypotenuse = x

side = y

(x+y) → constant

$\cos \theta=\frac{y}{x}$

area = $\frac{1}{2}\left(\sqrt{x^2-y^2}\right) \times y$        .....(1)

Let  x + y = k  ⇒  x = k - y

substituting in (1)

we get area = $\frac{1}{2} \sqrt{(k-y)^2-y^2} \times y$

area = $\frac{1}{2} \sqrt{k^2+y^2-y^2+2 k y} \times y$

area = $\frac{1}{2} \sqrt{k^2-2 k y} \times y$

squaring both sides

(area)² = $\frac{1}{4}\left(k^2-2 k y\right) y^2$         [Let area² = A]

$A=\frac{1}{4}\left(k^2 y^2-2 k y^3\right)$      ......(2)

for area maximum/minimum   $\frac{dy}{dx} = 0$

differentiating (2)

$\frac{1}{4}\left(2 k^2 y-6 k y^2\right)=0$

$2 k^2 y=6 k y^2$

$k y=3 y^2$

$k=3 y$

$x+y=3 y$    (as k = x + y)

$x=2 y$

$\frac{x}{y}=2$

$\frac{1}{2} =\frac{y}{x}$

$\Rightarrow \cos \theta =\frac{1}{2}$           (as $\frac{y}{x} = \cos \theta$)

$\theta=\cos ^{-1}(\frac{1}{2})=\frac{\pi}{3}$