Question

In $△ABC,{\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2,$ then the triangle is always

• A. an isosceles triangle
• B. a right angle triangle
• C. an obtuse angle triangle
• D. an acute angle triangle

Answer 1

The correct option is B a right angle triangle

${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2\Rightarrow \frac{1-\cos 2A}{2}+\frac{1-\cos 2B}{2}+\frac{1-\cos 2C}{2}=2\Rightarrow \cos 2A+\cos 2B+\cos 2C=-1\Rightarrow 2\cos (A+B)\cos (A-B)=-2{\cos }^{2}C\Rightarrow 2\cos (\pi -C)\cos (A-B)=-2{\cos }^{2}C[\because A+B+C=\pi ]\Rightarrow 2\cos C[\cos C-\cos (A-B\left)\right]=0[\because \cos (\pi -C)=-\cos C]\Rightarrow -2\cos C\left[\cos \right(A+B)+\cos (A-B\left)\right]=0[\because \cos (\pi -(A+B))=-\cos C]\Rightarrow -4\cos A\cos B\cos C=0\Rightarrow \cos A\cos B\cos C=0$
Therefore, one of them should be equal to zero.
So, exactly one of angle is ${90}^{\circ }$

Hence, the triangle is always a right angle triangle.