Let $a, b$ and $c$ be the lengths of the sides of a triangle with its opposite angles $A, B$ and $C$ respectively. If $\angle C=60^{\circ}$, then the value of $\frac{c(a+b)+\left(a^2+b^2\right)}{(b+c)(c+a)}$ is

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Aicharya · Accepted Answer

$\begin{aligned} & \cos C=\frac{a^2+b^2-c^2}{2 a b} \ & \Rightarrow \quad \cos 60^{\circ}=\frac{a^2+b^2-c^2}{2 a b} \ & \Rightarrow a^2+b^2-c^2=a b \ & \Rightarrow \quad a^2+b^2=a b+c^2 \ & \text {Now, } \frac{c(a+b)+\left(a^2+b^2\right)}{(b+c)(c+a)}=\frac{c a+b c+a b+c^2}{b c+a b+c^2+a c}=1 \end{aligned}$

Let \(a, b\) and \(c\) be the lengths of the sides of a triangle with its opposite angles \(A, B\) and \(C\)…

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