Question

In \(\triangle ABC\), with usual notations, if \(a^4 + b^4 + c^4 - 2a^2 c^2 - 2c^2 b^2 = 0\), then \(\angle C = \dots\)

• A. \(135^\circ\)
• B. \(120^\circ\)
• C. \(150^\circ\)
• D. \(125^\circ\)

Hint:

In triangle questions with \(a,b,c\), try converting the expression into \[ a^2+b^2-c^2 \] because it connects directly with \(\cos C\) using the cosine rule.

Answer 1

The Correct Option is A

Concept:
In a triangle, the cosine rule gives: \[ c^2 = a^2 + b^2 - 2ab\cos C \] So if we can reduce the given expression to a form involving \[ a^2+b^2-c^2 \] then we can directly use the cosine rule. ip

Step 1: Rewrite the given expression carefully.
Given: \[ a^4+b^4+c^4-2a^2c^2-2b^2c^2=0 \] Now write \[ a^4+b^4 = (a^2+b^2)^2 - 2a^2b^2 \] So the given equation becomes: \[ (a^2+b^2)^2 - 2a^2b^2 + c^4 - 2c^2(a^2+b^2)=0 \] Hence, \[ (a^2+b^2-c^2)^2 - 2a^2b^2 = 0 \] ip

Step 2: Use cosine rule.
From cosine rule, \[ a^2+b^2-c^2 = 2ab\cos C \] Substitute into the transformed equation: \[ (2ab\cos C)^2 - 2a^2b^2 = 0 \] \[ 4a^2b^2\cos^2 C - 2a^2b^2 = 0 \] ip

Step 3: Solve for \(\cos C\).
\[ 2a^2b^2(2\cos^2 C - 1)=0 \] Since \(a,b\neq 0\), we get \[ 2\cos^2 C - 1 = 0 \] \[ \cos^2 C = \frac{1}{2} \] \[ \cos C = \pm \frac{1}{\sqrt{2}} \] So, \[ C = 45^\circ \text{ or } 135^\circ \] ip

Step 4: Match with the given options.
Among the given options, only \[ 135^\circ \] is present. ip Hence, the correct answer is:
\[ \boxed{(A)\ 135^\circ} \]