If the sides of a triangle are 3, 5, 7,then | vedclass

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(B) Let the sides be $a = 3$,$b = 5$,and $c = 7$.To determine the nature of the triangle,we calculate the cosine of the largest angle,which is opposit

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One angle is obtuse

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(B) Let the sides be $a = 3$,$b = 5$,and $c = 7$.To determine the nature of the triangle,we calculate the cosine of the largest angle,which is opposite the longest side $(c = 7)$.Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.Substituting the values: $\cos C = \frac{3^2 + 5^2 - 7^2}{2 	imes 3 	imes 5} = \frac{9 + 25 - 49}{30} = \frac{34 - 49}{30} = \frac{-15}{30} = -\frac{1}{2}$.Since $\cos C = -\frac{1}{2}$,the angle $C = 120^{\circ}$.Because $C > 90^{\circ}$,the triangle has one obtuse angle.

If the sides of a triangle are $3, 5, 7$,then

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