In triangle ABC,with usual notations,if a^4+b | vedclass

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(A) Given the equation: $a^4+b^4+c^4-2a^2c^2-2c^2b^2=0$. Rearranging the terms,we have: $a^4+b^4+c^4-2a^2c^2-2b^2c^2 = 0$. This can be written as: $a^

Vedclass · Accepted Answer

$135$

Vedclass · Answer

(A) Given the equation: $a^4+b^4+c^4-2a^2c^2-2c^2b^2=0$. Rearranging the terms,we have: $a^4+b^4+c^4-2a^2c^2-2b^2c^2 = 0$. This can be written as: $a^4+b^4+c^4-2a^2c^2-2b^2c^2+2a^2b^2-2a^2b^2 = 0$. $(a^2+b^2-c^2)^2 - 2a^2b^2 = 0$. $(a^2+b^2-c^2)^2 = 2a^2b^2$. Taking the square root on both sides: $a^2+b^2-c^2 = \pm \sqrt{2}ab$. Using the Law of Cosines: $\cos C = \frac{a^2+b^2-c^2}{2ab}$. Substituting the expression: $\cos C = \frac{\pm \sqrt{2}ab}{2ab} = \pm \frac{\sqrt{2}}{2} = \pm \frac{1}{\sqrt{2}}$. Since $\angle C$ is an angle in a triangle,$\cos C$ can be $\frac{1}{\sqrt{2}}$ or $-\frac{1}{\sqrt{2}}$. If $\cos C = \frac{1}{\sqrt{2}}$,then $C = 45^{\circ}$. If $\cos C = -\frac{1}{\sqrt{2}}$,then $C = 135^{\circ}$. Given the options,$135^{\circ}$ is the correct choice.

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

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