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 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

The correct option is C $4$

Given Condition:$2{\left(2R\right)}^2=\frac{1}{2}(a^2+b^2+c^2)$
$\Rightarrow a^2+b^2+c^2=16R^2$
Using sine rule, we get
$\Rightarrow {\left(2R\sin A\right)}^2+{\left(2R\sin B\right)}^2+{\left(2R\sin C\right)}^2=16R^2$
$\Rightarrow 4R^2[{\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C]=16R^2$
$\Rightarrow {\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=4$