Question

If \[a,b,\]and \[c\]are the sides of a triangle such that \[{a^4} + {b^4} + {c^4} = 2{c^2}({a^2} + {b^2})\] then find the angle opposite to side \[c\].
A. \[{45^o}\]or \[{135^o}\]
B. \[{30^o}\]or \[{100^o}\]
C. \[{50^o}\]or \[{100^o}\]
D. \[{60^o}\]or \[{120^o}\]

Answer 1

Hint:
In the given question, we need to find the angle opposite to side \[c\]. For this, we will simplify the given equation and also use the trigonometric identities to get the desired result.



Formula Used:
In \[ \Delta ABC \],
By cosine rule: \[\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}\]
\[(a+b+c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca\]



Complete step-by-step answer:
Consider the following figure.

In \[ \Delta ABC \], angle \[C\] is the opposite angle to side \[c\]

Given expression: \[{a^4} + {b^4} + {c^4} = 2{c^2}({a^2} + {b^2})\]
Now, we will simplify the expression,
\[{a^4} + {b^4} + {c^4} = 2{c^2}{a^2} + 2{c^2}{b^2}\]
\[{a^4} + {b^4} + {c^4} - 2{c^2}{a^2} - 2{c^2}{b^2} = 0\]
By adding \[2{a^2}{b^2}\] on both sides, we get;
\[{a^4} + {b^4} + {c^4} - 2{c^2}{a^2} - 2{c^2}{b^2} + 2{a^2}{b^2} = 2{a^2}{b^2}\]
\[{\left( {{a^2} + {b^2} - {c^2}} \right)^2} = 2{\left( {ab} \right)^2}\]
Now, by taking square root on both sides, we get
\[\left( {{a^2} + {b^2} - {c^2}} \right) = \pm \sqrt 2 \left( {ab} \right)\]
Divide by \[2\] on both sides.
\[\dfrac{{\left( {{a^2} + {b^2} - {c^2}} \right)}}{2} = \dfrac{{ \pm \sqrt 2 \left( {ab} \right)}}{2}\]
By simplifying, we get;
\[\dfrac{{\left( {{a^2} + {b^2} - {c^2}} \right)}}{{2ab}} = \dfrac{{ \pm \sqrt 2 }}{2}\]
Since, by cosine rule: \[\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}\]
\[\Rightarrow \cos C = \dfrac{{ \pm 1}}{{\sqrt 2 }}\]
Therefore, \[\angle C = {45^o}\] or \[\angle C = {135^o}\]

Hence, the correct option is (A).



Note:
Many students make mistakes in the calculation part as well as writing trigonometric identity. This is the only way through which we can solve the example in the simplest way. Also, it is essential to get the correct angle \[C\].