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In triangle $\mathrm{ABC}$, if $\frac{\sin ^{2} \mathrm{~A}+\sin ^{2} \mathrm{~B}+\sin ^{2} \mathrm{C}}{\cos ^{2} \mathrm{~A}+\cos ^{2} \mathrm{~B}+\cos ^{2} \mathrm{C}}=2$ then the
triangle is
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triangle is
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Verified Answer
The correct answer is:
right-angled
Given, $\frac{\sin ^{2} A+\sin ^{2} B+\sin ^{2} C}{\cos ^{2} A+\cos ^{2} B+\cos ^{2} C}=2$.
Let us take $\mathrm{A}=30^{\circ}, \mathrm{B}=60^{\circ}, \mathrm{C}=90^{\circ}$
$\frac{\sin ^{2} 30^{\circ}+\sin ^{2} 60^{\circ}+\sin ^{2} 90^{\circ}}{\cos ^{2} 30^{\circ}+\cos ^{2} 60^{\circ}+\cos ^{2} 90^{\circ}}$
$=\frac{\frac{1}{4}+\frac{3}{4}+1}{\frac{3}{4}+\frac{1}{4}+0}=\frac{1+1}{1}=\frac{2}{1}=2$
So, the given triangle is right-angled triangle.
Let us take $\mathrm{A}=30^{\circ}, \mathrm{B}=60^{\circ}, \mathrm{C}=90^{\circ}$
$\frac{\sin ^{2} 30^{\circ}+\sin ^{2} 60^{\circ}+\sin ^{2} 90^{\circ}}{\cos ^{2} 30^{\circ}+\cos ^{2} 60^{\circ}+\cos ^{2} 90^{\circ}}$
$=\frac{\frac{1}{4}+\frac{3}{4}+1}{\frac{3}{4}+\frac{1}{4}+0}=\frac{1+1}{1}=\frac{2}{1}=2$
So, the given triangle is right-angled triangle.
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