Maximizing Triangle Area

To find the angle that maximizes the area of a right-angled triangle when the sum of the hypotenuse and one side is constant, let's denote the sides of the triangle.

• Let the sides of the right-angled triangle be \(a\), \(b\), and the hypotenuse be \(c\).
• The area of the triangle, \(A\), is given by \(A = \frac{1}{2}ab\).
• According to the Pythagorean theorem, \(a^2 + b^2 = c^2\).
• The problem states that the sum of the lengths of the hypotenuse and a side is constant. Let's assume this side is \(b\). So, \(b + c = k\), where \(k\) is a constant.

Expressing Area in Terms of One Variable

From the constant sum, we can express \(c\) in terms of \(k\) and \(b\):

\[ c = k - b \]

Now, substitute this into the Pythagorean theorem to find \(a\) in terms of \(k\) and \(b\):

\[ a^2 + b^2 = (k - b)^2 \]

Expand the right side:

\[ a^2 + b^2 = k^2 - 2kb + b^2 \]

Subtract \(b^2\) from both sides:

\[ a^2 = k^2 - 2kb \]

So, the side \(a\) is:

\[ a = \sqrt{k^2 - 2kb} \]

Now, substitute \(a\) into the area formula:

\[ A = \frac{1}{2} b \sqrt{k^2 - 2kb} \]

Maximizing Area Using Differentiation

To maximize \(A\), it is equivalent to maximize \(A^2\), which simplifies the differentiation process:

\[ A^2 = \left( \frac{1}{2} b \sqrt{k^2 - 2kb} \right)^2 \]

\[ A^2 = \frac{1}{4} b^2 (k^2 - 2kb) \]

\[ A^2 = \frac{1}{4} (k^2 b^2 - 2kb^3) \]

Let \(f(b) = k^2 b^2 - 2kb^3\). To find the maximum value of \(f(b)\), we need to find its derivative with respect to \(b\) and set it to zero:

\[ f'(b) = \frac{d}{db} (k^2 b^2 - 2kb^3) \]

\[ f'(b) = 2k^2 b - 6kb^2 \]

Set \(f'(b) = 0\) to find critical points:

\[ 2k^2 b - 6kb^2 = 0 \]

Factor out \(2kb\):

\[ 2kb (k - 3b) = 0 \]

This gives two possible solutions: \(2kb = 0\) or \(k - 3b = 0\).

• Since \(b\) is a length, \(b \neq 0\).
• Therefore, \(k - 3b = 0\).
• Solving for \(b\), we get \(b = \frac{k}{3}\).

Now we find the corresponding value of \(c\):

\[ c = k - b = k - \frac{k}{3} = \frac{3k - k}{3} = \frac{2k}{3} \]

So, for maximum area, the side \(b = \frac{k}{3}\) and the hypotenuse \(c = \frac{2k}{3}\).

Calculating the Angle

The question asks for the angle between the hypotenuse (\(c\)) and the side (\(b\)). In a right-angled triangle, the cosine of an angle is the ratio of the adjacent side to the hypotenuse.

Let \(\theta\) be the angle between the hypotenuse \(c\) and the side \(b\).

\[ \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{b}{c} \]

Substitute the values of \(b\) and \(c\) we found:

\[ \cos(\theta) = \frac{\frac{k}{3}}{\frac{2k}{3}} \]

\[ \cos(\theta) = \frac{k}{3} \cdot \frac{3}{2k} \]

\[ \cos(\theta) = \frac{1}{2} \]

To find \(\theta\), we take the inverse cosine:

\[ \theta = \arccos\left(\frac{1}{2}\right) \]

\[ \theta = 60^\circ \]

Thus, for the area of the right-angled triangle to be maximum, the angle between the hypotenuse and the side (whose sum with hypotenuse is constant) must be \(60^\circ\).

The final answer is 60°