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(C) Let the sides of the triangle be $a = 3$, $b = 5$, and $c = 7$.The largest angle is opposite the longest side, which is $c = 7$. Let this angle be

Vedclass · Accepted Answer

$\frac{2\pi}{3}$

Vedclass · Answer

(C) Let the sides of the triangle be $a = 3$, $b = 5$, and $c = 7$.The largest angle is opposite the longest side, which is $c = 7$. Let this angle be $C$.Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.Substitute the values: $\cos C = \frac{3^2 + 5^2 - 7^2}{2(3)(5)}$.$\cos C = \frac{9 + 25 - 49}{30} = \frac{34 - 49}{30} = \frac{-15}{30} = -\frac{1}{2}$.Since $\cos C = -\frac{1}{2}$, the angle $C = \arccos(-\frac{1}{2}) = \frac{2\pi}{3}$.

With the usual notations, if the lengths of the sides of a triangle are $3 \text{ units}$, $5 \text{ units}$, and $7 \text{ units}$, then the largest angle of the triangle is

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