In triangle ABC, angle B=90^{circ} and (b+a) | vedclass

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(C) Establish Relationships:In $	riangle ABC$, we are given $\angle B=90^{\circ}$, making it a right-angled triangle where $b$ is the hypotenuse and

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$\frac{\pi}{3}$

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(C) Establish Relationships:In $	riangle ABC$, we are given $\angle B=90^{\circ}$, making it a right-angled triangle where $b$ is the hypotenuse and $a, c$ are the legs. From the Pythagorean theorem, $c=\sqrt{b^2-a^2}$. We are given that $a+b=k$ (a constant), so $b=k-a$. The area $A$ is:$A=\frac{1}{2}ac=\frac{1}{2}a\sqrt{b^2-a^2}$Express Area in Terms of a Single Variable:Substitute $b=k-a$ into the area equation:$A=\frac{1}{2}a\sqrt{(k-a)^2-a^2}=\frac{1}{2}a\sqrt{k^2-2ka+a^2-a^2}=\frac{1}{2}a\sqrt{k^2-2ka}$To maximize $A$, we maximize $f(a)=4A^2=a^2(k^2-2ka)=k^2a^2-2ka^3$.Differentiate and Solve for Extremum:Find the derivative of $f(a)$ with respect to $a$ and set it to zero:$f'(a)=2k^2a-6ka^2=0 \Longrightarrow 2ka(k-3a)=0$Since $a, k 
eq 0$, we have $a=\frac{k}{3}$.Then $b=k-\frac{k}{3}=\frac{2k}{3}$.Determine Angle $C$:In the right triangle, $\cos C=\frac{a}{b}$.$\cos C=\frac{a}{b}=\frac{k/3}{2k/3}=\frac{1}{2}$Since $\cos C=\frac{1}{2}$, then $C=\frac{\pi}{3}$.

In $\triangle ABC$, $\angle B=90^{\circ}$ and $(b+a)$ is always a constant. In order that $\triangle ABC$ encloses the maximum area, $\angle C=$

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