Question

In $\Delta ABC$ if $si{n}^{2}A+si{n}^{2}B+si{n}^{2}C=2,$ then the triangle is

• A. right angled, but need not be isosceles
• B. right angled and isosceles
• C. isosceles, but need not be right angled
• D. equilateral

Answer 1

Answer: (A)

Given, ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2$
$\Rightarrow$ $1-{\cos }^{2}A+1-{\cos }^{2}B+1-{\cos }^{2}C=2$
$\Rightarrow$ $1={\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C$
$\Rightarrow$ $1=1-2\cos A\cos B\cos C$
$\Rightarrow$ $\cos A\cos B\cos C=0$
At least one angle should be ${90}^o$ and sum of two angles should be ${90}^o$ .
Hence, option [a] is correct.