If a=3, b=5, c=7 are the sides of a triangle | vedclass

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(A) The sides of the triangle are $a=3, b=5, c=7$. First,we find the cosine of angle $C$ using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2a

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$\frac{7}{\sqrt{3}}$

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(A) The sides of the triangle are $a=3, b=5, c=7$. First,we find the cosine of angle $C$ using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{3^2 + 5^2 - 7^2}{2(3)(5)} = \frac{9 + 25 - 49}{30} = \frac{-15}{30} = -\frac{1}{2}$. Since $\cos C = -\frac{1}{2}$,we have $C = 120^\circ$. The area of the triangle $\Delta$ is given by $\Delta = \frac{1}{2}ab \sin C = \frac{1}{2}(3)(5) \sin(120^\circ) = \frac{15}{2} 	imes \frac{\sqrt{3}}{2} = \frac{15\sqrt{3}}{4}$. The circumradius $R$ is given by $R = \frac{abc}{4\Delta} = \frac{3 	imes 5 	imes 7}{4 	imes (\frac{15\sqrt{3}}{4})} = \frac{105}{15\sqrt{3}} = \frac{7}{\sqrt{3}}$.

If $a=3, b=5, c=7$ are the sides of a triangle $ABC$,then its circumradius is

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