If a = 2, b = 3, c = 5 in Delta ABC,then C = | vedclass

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(D) Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.Substituting the given values: $\cos C = \frac{2^2 + 3^2 - 5^2}{2(2)(3)}$.$\cos C

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(D) Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.Substituting the given values: $\cos C = \frac{2^2 + 3^2 - 5^2}{2(2)(3)}$.$\cos C = \frac{4 + 9 - 25}{12} = \frac{13 - 25}{12} = \frac{-12}{12} = -1$.Since $\cos C = -1$,we have $C = 180^{\circ}$ or $\pi$ radians.However,the sum of two sides of a triangle must be greater than the third side $(a + b > c)$.Here,$2 + 3 = 5$,which means $a + b = c$.This implies that the points $A, B, C$ are collinear and do not form a triangle.Therefore,the correct option is $D$.

If $a = 2, b = 3, c = 5$ in $\Delta ABC$,then $C = $

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