In triangle ABC,if sin^2 A + sin^2 B = sin^2 | vedclass

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(C) Given $\sin^2 A + \sin^2 B = \sin^2 C$. Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $\sin A = \frac{

Vedclass · Accepted Answer

$25$

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(C) Given $\sin^2 A + \sin^2 B = \sin^2 C$. Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $\sin A = \frac{a}{2R}$,$\sin B = \frac{b}{2R}$,and $\sin C = \frac{c}{2R}$. Substituting these into the given equation: $(\frac{a}{2R})^2 + (\frac{b}{2R})^2 = (\frac{c}{2R})^2 \Rightarrow a^2 + b^2 = c^2$. This implies that $	riangle ABC$ is a right-angled triangle with the hypotenuse $c = l(AB) = 10$. The area of the triangle is $Area = \frac{1}{2}ab$. Since $c = 10$,we have $a = 10 \sin A$ and $b = 10 \cos A$. $Area = \frac{1}{2}(10 \sin A)(10 \cos A) = 50 \sin A \cos A = 25 \sin(2A)$. The maximum value of $\sin(2A)$ is $1$ (when $2A = 90^{\circ}$ or $A = 45^{\circ}$). Therefore,the maximum area is $25 	imes 1 = 25$.

In $\triangle ABC$,if $\sin^2 A + \sin^2 B = \sin^2 C$ and $l(AB) = 10$,then the maximum value of the area of $\triangle ABC$ is

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