Search any question & find its solution
Question:
Answered & Verified by Expert
In a $\triangle A B C, a^4+b^4+c^4=2 b^2 c^2+2 a^2 b^2$, then $B=$
Options:
Solution:
1318 Upvotes
Verified Answer
The correct answer is:
$\frac{\pi}{4}$ or $\frac{3 \pi}{4}$
We have, $\quad a^4+b^4+c^4=2 b^2\left(c^2+a^2\right)$
$$
\Rightarrow a^4+b^4+c^4-2 b^2 c^2-2 a^2 b^2=0
$$
Add $2 a^2 c^2$ on both sides, we get
$$
\begin{array}{ll}
\Rightarrow a^4+b^4+c^4-2 b^2 c^2-2 a^2 b^2+2 c^2 a^2=2 c^2 a^2 \\
\Rightarrow \quad\left(a^2+c^2-b^2\right)^2=2 c^2 a^2 \\
\Rightarrow & \frac{4\left(a^2+c^2-b^2\right)^2}{(2 c a)^2}=2 \\
\Rightarrow & \frac{\left(a^2+c^2-b^2\right)^2}{(2 c a)^2}=\frac{1}{2} \quad\left[\because \cos B=\frac{c^2+a^2-b^2}{2 c a}\right] \\
\therefore & \cos ^2 B=\frac{1}{2} \\
\Rightarrow & \cos B=\frac{1}{\sqrt{2}}
\end{array}
$$
So, we get $\quad \angle C=\frac{\pi}{4}$ or $\frac{3 \pi}{4}$
$$
\Rightarrow a^4+b^4+c^4-2 b^2 c^2-2 a^2 b^2=0
$$
Add $2 a^2 c^2$ on both sides, we get
$$
\begin{array}{ll}
\Rightarrow a^4+b^4+c^4-2 b^2 c^2-2 a^2 b^2+2 c^2 a^2=2 c^2 a^2 \\
\Rightarrow \quad\left(a^2+c^2-b^2\right)^2=2 c^2 a^2 \\
\Rightarrow & \frac{4\left(a^2+c^2-b^2\right)^2}{(2 c a)^2}=2 \\
\Rightarrow & \frac{\left(a^2+c^2-b^2\right)^2}{(2 c a)^2}=\frac{1}{2} \quad\left[\because \cos B=\frac{c^2+a^2-b^2}{2 c a}\right] \\
\therefore & \cos ^2 B=\frac{1}{2} \\
\Rightarrow & \cos B=\frac{1}{\sqrt{2}}
\end{array}
$$
So, we get $\quad \angle C=\frac{\pi}{4}$ or $\frac{3 \pi}{4}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.