AP EAMCET · Maths · Properties of Triangles
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
- A \(\frac{1}{2}\)
- B \(\frac{\sqrt{3}}{2}\)
- C 1
- D \(\sqrt{3}\)
Answer & Solution
Correct Answer
(C) 1
Step-by-step Solution
Detailed explanation
\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…
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