Question

If twice the square of the diameter of a circle is equal to half the sum of the squares of the sides of inscribed $\Delta$ABC,

then $si{n}^{2}A+si{n}^{2}B+si{n}^{2}C$ is equal to?

• A. $1$
• B. $2$
• C. $4$
• D. $8$

Answer 1

The correct option is C $4$

$2{\left(2R\right)}^2=\frac{1}{2}.a^2+b^2+c^2=\frac{{\left(2R\right)}^2}{2}({\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C)\Rightarrow {\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=4$