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

Step 1: Understanding the Concept:
We use the Cosine Rule: \(\cos C = \frac{a^2 + b^2 - c^2}{2ab}\).
Step 2: Key Formula or Approach:
The given equation can be rewritten as a squared term.
Observe: \(x^2 + y^2 + z^2 - 2xz - 2yz + 2xy = (x + y - z)^2\).
Step 3: Detailed Explanation:
\(a^4 + b^4 + c^4 - 2a^2 c^2 - 2b^2 c^2 = 0\)
Add \(2a^2 b^2\) to both sides:
\(a^4 + b^4 + c^4 - 2a^2 c^2 - 2b^2 c^2 + 2a^2 b^2 = 2a^2 b^2\)
\((a^2 + b^2 - c^2)^2 = 2a^2 b^2\)
Taking square root: \(a^2 + b^2 - c^2 = \pm \sqrt{2} ab\)
Now, \(\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{-\sqrt{2} ab}{2ab} = \frac{-1}{\sqrt{2}}\). (The positive root gives \(45^\circ\), not in options).
If \(\cos C = \frac{-1}{\sqrt{2}}\), then \(C = 135^\circ\).
Step 4: Final Answer:
The angle \(\angle C\) is \(135^\circ\).