Question

In a triangle $ABC$, $c^4-2c^2\left(a^2+b^2\right)+a^4+a^2{b}^2+b^4=0$, then the acute angle $C$ is equal to

• A. $30°$
• B. $60°$
• C. $45°$
• D. $75°$

Answer 1

The correct option is B

$60°$

Explanation for the correct option.

Step 1. Simplify the given equation.

The equation $c^4-2c^2\left(a^2+b^2\right)+a^4+a^2{b}^2+b^4=0$ can be written as:

${\left(a^2\right)}^2+{\left(b^2\right)}^2+{\left(-c^2\right)}^2+a^2{b}^2+2\left(a^2\right)\left(-c^2\right)+2\left(b^2\right)\left(-c^2\right)=0$

Now add $a^2{b}^2$ both sides and form the perfect square.

$$\begin{gathered} {\left(a^2\right)}^2+{\left(b^2\right)}^2+{\left(-c^2\right)}^2+a^2{b}^2+2\left(a^2\right)\left(-c^2\right)+2\left(b^2\right)\left(-c^2\right)+a^2{b}^2=a^2{b}^2 \\ \Rightarrow {\left(a^2\right)}^2+{\left(b^2\right)}^2+{\left(-c^2\right)}^2+2a^2{b}^2+2\left(a^2\right)\left(-c^2\right)+2\left(b^2\right)\left(-c^2\right)=a^2{b}^2 \\ \Rightarrow {\left(a^2+b^2-c^2\right)}^2={\left(ab\right)}^2 \\ \Rightarrow a^2+b^2-c^2=ab \end{gathered}$$

Step 2. Find the acute angle $C$.

Using the cosine rule it is known that $\cos C=\frac{a^2+b^2-c^2}{2ab}$ and as $a^2+b^2-c^2=ab$, so

$$\begin{gathered} \cos C=\frac{ab}{2ab} \\ \Rightarrow \cos C=\frac{1}{2} \\ \Rightarrow C=60° \end{gathered}$$

So the acute angle $C$ is equal to $60°$.

Hence the correct option is B.