Question

If in a $△ABC,a^4+b^4+c^4=2a^2{b}^2+b^2{c}^2+2c^2{a}^2,$ then $\sin A$ is equal to

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{\sqrt{3}+1}{2\sqrt{2}}$

Answer 1

Answer: (B)

$\frac{1}{2}$

$\because \cos A=\frac{b^2+c^2-a^2}{2bc}$
$\Rightarrow b^2+c^2-a^2=2bc\cos A$
Squaring both sides, we get
$\Rightarrow {(b^2+c^2-a^2)}^2=4b^2{c}^2{\cos }^{2}A$
$b^4+c^4+a^4+2b^2{c}^2-2c^2{a}^2-2a^2{b}^2=4b^2{c}^2{\cos }^{2}A$
Given: $a^4+b^4+c^4=2a^2{b}^2+b^2{c}^2+2c^2{a}^2$
$\Rightarrow (a^4+b^4+c^4)+2b^2{c}^2-2c^2{a}^2-2a^2{b}^2=4b^2{c}^2{\cos }^{2}A$
$\Rightarrow (2a^2{b}^2+b^2{c}^2+2c^2{a}^2)+2b^2{c}^2-2c^2{a}^2-2a^2{b}^2=4b^2{c}^2{\cos }^{2}A$
$\Rightarrow 3b^2{c}^2=4b^2{c}^2{\cos }^{2}A$
$\Rightarrow 3=4{\cos }^{2}A$
$\Rightarrow {\cos }^{2}A=\frac{3}{4}$
$\Rightarrow 1-{\sin }^{2}A=\frac{3}{4}$
$\Rightarrow {\sin }^{2}A=1-\frac{3}{4}=\frac{1}{4}$
$\therefore \sin A=\pm \frac{1}{2}$