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Q.

In $△ ABC, if c^{4} - 2 (a^{2} + b^{2}) c^{2} + a^{4} + a^{2} b^{2} + b^{4} = 0 then ∠ C =$

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a

30°

b

45°

c

60°

d

75°

answer is C.

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Detailed Solution

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${(c^{2})}^{2} + {(a^{2})}^{2} + {(b^{2})}^{2} - 2 a^{2} c^{2} - 2 b^{2} c^{2} + 2 a^{2} b^{2} = a^{2} b^{2} \Rightarrow {(a^{2} + b^{2} - c^{2})}^{2} = a^{2} b^{2} \Rightarrow a^{2} + b^{2} - c^{2} = \pm ab \Rightarrow \frac{a^{2} + b^{2} - c^{2}}{2 ab} = \pm \frac{1}{2} \Rightarrow cosC = \pm \frac{1}{2} ∴ C = 60 ° or 120 °$

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