Question

In a triangle $\vartriangle ABC$, if $2{{a}^{2}}{{b}^{2}}+2{{b}^{2}}{{c}^{2}}={{a}^{4}}+{{b}^{4}}+{{c}^{4}}$, then angle $B$ is equal to
A. ${{45}^{0}}$ or ${{135}^{0}}$
B. ${{135}^{0}}$or ${{120}^{0}}$
C. ${{30}^{0}}$or ${{60}^{0}}$
D. None of these.

Answer 1

Hint: We will use the equation given by writing it in the form of expansion of \[{{(a-b+c)}^{2}}\] by adding suitable terms in the equation on both the sides to derive the value of \[{{a}^{2}}-{{b}^{2}}+{{c}^{2}}\]. Then using cosine rule ${{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\cos B$ we will determine the value of the required angle.

Formula Used: ${{(a-b+c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}-2ab-2bc+2ac$

Complete step by step solution: We are given a $\vartriangle ABC$ such that $2{{a}^{2}}{{b}^{2}}+2{{b}^{2}}{{c}^{2}}={{a}^{4}}+{{b}^{4}}+{{c}^{4}}$ and we have to find the value of angle$~B$.
We will take the given equation,
\[{{a}^{4}}+{{b}^{4}}+{{c}^{4}}-2{{a}^{2}}{{b}^{2}}-2{{b}^{2}}{{c}^{2}}=0\]
We will now add \[2{{a}^{2}}{{c}^{2}}\] on both the sides of the equation.
\[{{a}^{4}}+{{b}^{4}}+{{c}^{4}}-2{{b}^{2}}{{a}^{2}}-2{{b}^{2}}{{c}^{2}}+2{{a}^{2}}{{c}^{2}}=2{{a}^{2}}{{c}^{2}}\]
Now we can write this equation as,
\[\begin{align}
  & {{({{a}^{2}}-{{b}^{2}}+{{c}^{2}})}^{2}}=2{{a}^{2}}{{c}^{2}} \\
 & {{a}^{2}}-{{b}^{2}}+{{c}^{2}}=\sqrt{2{{a}^{2}}{{c}^{2}}} \\
 & {{a}^{2}}-{{b}^{2}}+{{c}^{2}}=\pm \sqrt{2}ac
\end{align}\]
We will now use cosine rule ${{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\cos B$,
\[\begin{align}
  & 2ac\cos B={{a}^{2}}-{{b}^{2}}+{{c}^{2}} \\
 & \cos B=\dfrac{{{a}^{2}}-{{b}^{2}}+{{c}^{2}}}{2ac}
\end{align}\]
Now we will substitute the value of the equation \[{{a}^{2}}-{{b}^{2}}+{{c}^{2}}=\pm \sqrt{2}ac\] we derived in above equation.
\[\cos B=\dfrac{\pm \sqrt{2}ac}{2ac}\]
\[\begin{align}
  & \cos B=\dfrac{\pm \sqrt{2}}{2} \\
 & \cos B=\dfrac{1}{\sqrt{2}}\,\,\,or\,\,\dfrac{-1}{\sqrt{2}} \\
 & \cos B=\cos {{45}^{0}}\,\,or\,\,\cos {{135}^{0}} \\
\end{align}\]
Hence the value of angle will be,
\[B={{45}^{0}}\,\,or\,\,{{135}^{0}}\]

The angle $B$ of the triangle $\vartriangle ABC$ such that $2{{a}^{2}}{{b}^{2}}+2{{b}^{2}}{{c}^{2}}={{a}^{4}}+{{b}^{4}}+{{c}^{4}}$is \[B={{45}^{0}}\,\,or\,\,{{135}^{0}}\]. Hence the correct option is (A).

Note: The Law of cosines shows the relationship between length of the sides of a triangle with respect to the cosine of its angle.
The Law of cosines is also known as cosine rule does not holds true for all the triangles. It can be also used when all the three sides of the triangle are given and we have to determine the angles of the triangle.
There are three laws of cosine:
${{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos A$
${{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\cos B$
${{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos c$
Here $a,b$ and $c$ are the sides of the triangle and $A$ is the angle opposite to side, $B$ is the angle opposite to side $b$ and $C$ is the angle opposite to side $c$.