MCQ
$\triangle ABC$ માટે $a+b+c$
$0$ હોય અને $ \begin{vmatrix}\mathbf{a} & \mathbf{b} & \mathbf{c} \\ b & c & a \\ c & a & b \end{vmatrix}=0$ હોય તો $sin^2A+sin^2B+sin^2C= ............... $
- A3.23
- B2.23
- C2.54
- ✓2.25
$D=3abc-a^3-b^3-c^3={0}$
$\therefore a^3+b^3+c^3-3abc={0}$
$\therefore \frac{1}{2}(a+b+c)[(a-b)^2+(b-c)^2-(c-a)^2]=0$
$a+b+c$
$0$ હોવાથી $a=b=c\Rightarrow A=B=C=\frac{\pi}{3}$
$sin^2A+sin^2B+sin^2C=\frac{3}{4}+\frac{3}{4}+\frac{3}{4}=2.25$
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