In a ∆ABC, if $\sin ^{2} \mathrm{~A}+\sin ^{2} \mathrm{~B}=\sin ^{2} \mathrm{C}$ show that the triangle is rightangled
Given: $\sin ^{2} \mathrm{~A}+\sin ^{2} \mathrm{~B}=\sin ^{2} \mathrm{C}$
Need to prove: The triangle is right-angled
$\sin ^{2} A+\sin ^{2} B=\sin ^{2} C$
We know, $\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}}=2 \mathrm{R}$
So,
$\sin ^{2} A+\sin ^{2} B=\sin ^{2} C$
$\frac{a^{2}}{4 R^{2}}+\frac{b^{2}}{4 R^{2}}=\frac{c^{2}}{4 R^{2}}$
$a^{2}+b^{2}=c^{2}$
This is one of the properties of right angled triangle. And it is satisfied here. Hence, the triangle is right angled. [Proved]
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