A) Right angled
B) Equilateral
C) Acute angled
D) Obtuse angled
Correct Answer: A
Solution :
\[8{{R}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}=4{{R}^{2}}({{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C)\] \[\Rightarrow \] \[{{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C=2\] \[\Rightarrow \] \[({{\cos }^{2}}A-{{\sin }^{2}}C)+{{\cos }^{2}}B=0\] \[\Rightarrow \] \[\cos (A-C)\cos (A+C)+{{\cos }^{2}}B=0\] Þ \[2\cos A\cos B\cos C=0\] So that, \[\cos A=0\] or \[\cos B=0\] or \[\cos C=0\] \[\Rightarrow \] \[A=\frac{\pi }{2}\] or \[B=\frac{\pi }{2}\] or \[C=\frac{\pi }{2}\].
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