Question

If twice the square of the diameter of a circle is equal to the sum of the squares on the sides of the inscribed triangle $ABC$,

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

• A. $1$
• B. $2$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $2$

we know that , $a=2RsinAb=2RsinBc=2RsinC$

Given: $2(2R{)}^2=a^2+b^2+c^2\Rightarrow 8R^2=4R^2(si{n}^{2}A+si{n}^{2}B+si{n}^{2}C)\Rightarrow si{n}^{2}A+si{n}^{2}B+si{n}^{2}C=2$

Therefore, Answer is $B$