Question

If the square of the diameter of a circle is equal to half the sum of the square of the sides on inscribed triangle $ABC$, then ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C$ is equal to

• A. $1$
• B. $2$
• C. $4$
• D. $8$

Answer 1

Answer: (B)

$2$

For a circle inscribed in a cricle, we have

$\frac{a}{2\sin A}=\frac{b}{2\sin B}=\frac{c}{2\sin C}=R$

${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=\frac{a^2}{4R^2}+\frac{b^2}{4R^2}+\frac{b^2}{4R^2}=\frac{(a^2+b^2+c^2)}{4R^2}$

It is given that $\frac{(a^2+b^2+c^2)}{2}={\left(2R\right)}^2\Rightarrow (a^2+b^2+c^2)=8R^2$

$\therefore {\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=\frac{8R^2}{4R^2}=2$