MHT CET · Maths · Properties of Triangles
In \(\triangle A B C\), with usual notations, if \(a^4+\mathrm{b}^4+\mathrm{c}^4-2 a^2 \mathrm{c}^2-2 \mathrm{c}^2 \mathrm{~b}^2=0\), then \(\angle \mathrm{C}=\ldots\).
- A \(135^{\circ}\)
- B \(120^{\circ}\)
- C \(150^{\circ}\)
- D \(125^{\circ}\)
Answer & Solution
Correct Answer
(A) \(135^{\circ}\)
Step-by-step Solution
Detailed explanation
\(c^4 - 2(a^2+b^2)c^2 + a^4+b^4 = 0\) \(c^2 = a^2+b^2 \pm \sqrt{(a^2+b^2)^2-(a^4+b^4)} = a^2+b^2 \pm \sqrt{2a^2b^2} = a^2+b^2 \pm \sqrt{2}ab\)
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