Question

If in a $△ABC$ ${\sin }^{4}A+{\sin }^{4}B+{\sin }^{4}C={\sin }^{2}B{\sin }^{2}C+2{\sin }^{2}C{\sin }^{2}A+2{\sin }^{2}A{\sin }^{2}B,$

then $A=$

• A. $\frac{\pi }{6},\frac{5\pi }{6}$
• B. $\frac{\pi }{3},\frac{5\pi }{6}$
• C. $\frac{\pi }{3},\frac{\pi }{2}$
• D. $\frac{\pi }{4},\frac{3\pi }{4}$

Answer 1

The correct option is A $\frac{\pi }{6},\frac{5\pi }{6}$

We know that
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ from the sine rule.

Hence, replacing the $\sin \theta$'s by the corresponding sides, we get
$a^4+b^4+c^4=b^2{c}^2+2a^2{c}^2+2a^2{b}^2$
$a^4+b^4+c^4-2a^2{c}^2-2a^2{b}^2-2b^2{c}^2=3b^2{c}^2$
$(b^2+c^2-a^2{)}^2=3b^2{c}^2\to (1)$
Now, $\cos A=\frac{b^2+c^2-a^2}{2bc}$
$\Rightarrow b^2+c^2-a^2=2bc\cos A$
Substituting in $\left(1\right)$,
$4b^2{c}^2{\cos }^{2}A=3b^2{c}^2$
${\cos }^{2}A=\frac{3}{4}$
$\cos A=\pm \frac{\sqrt{3}}{2}$
Hence,
$A=\frac{\pi }{6}$ and $A=\pi -\frac{\pi }{6}=\frac{5\pi }{6}$.