In triangle ABC,the sum of the lengths of two | vedclass

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Question

(A) Let the two sides be $a$ and $b$. We are given $a + b = x$ and $ab = y$. Given the condition $x^2 - c^2 = y$,we substitute $x = a + b$: $(a + b)^2

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$\frac{c}{\sqrt{3}}$

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(A) Let the two sides be $a$ and $b$. We are given $a + b = x$ and $ab = y$. Given the condition $x^2 - c^2 = y$,we substitute $x = a + b$: $(a + b)^2 - c^2 = y$ $a^2 + b^2 + 2ab - c^2 = y$ Since $ab = y$,we have $a^2 + b^2 + 2y - c^2 = y$,which simplifies to $a^2 + b^2 - c^2 = -y$. By the Law of Cosines,$c^2 = a^2 + b^2 - 2ab \cos C$. Comparing $a^2 + b^2 - c^2 = 2ab \cos C$ with $a^2 + b^2 - c^2 = -y$ and $ab = y$,we get $2y \cos C = -y$,so $\cos C = -\frac{1}{2}$. Thus,$C = 120^\circ$. The circumradius $R$ is given by $R = \frac{c}{2 \sin C}$. $R = \frac{c}{2 \sin 120^\circ} = \frac{c}{2 (\sqrt{3}/2)} = \frac{c}{\sqrt{3}}$.

In $\triangle ABC$,the sum of the lengths of two sides is $x$ and the product of those lengths is $y$. If $c$ is the length of the third side and $x^2 - c^2 = y$,then the circumradius of the triangle is

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