AP EAMCET · Maths · Properties of Triangles
\(\triangle A B C\), if \(a, b, c\) are its sides and \(\angle C=60^{\circ}\), find the value of \(\frac{a}{b+c}+\frac{b}{c+a}\)
- A 1
- B 0
- C \(\frac{\sqrt{3}}{2}\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(A) 1
Step-by-step Solution
Detailed explanation
In a \(\triangle A B C\), it is given \(\angle C=60^{\circ}\), then \[ \begin{aligned} & \cos 60^{\circ}=\frac{a^2+b^2-c^2}{2 a b} \\ & \Rightarrow a b=a^2+b^2-c^2 \Rightarrow c^2=a^2+b^2-a b \end{aligned} \] Now,…
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