Q67 / 160 · AP EAMCET 2022 (06 Jul Shift 2) · Mathematics

In a $\triangle A B C$, if $\angle A=60^{\circ}$, then $\frac{b}{c+a}+\frac{c}{a+b}=$

  • A
    $a+b+c$
  • B
    0
  • Correct answer
    C
    1
  • D
    $a b c$

Solution

Apply cosine rule, $\cos A=\frac{b^2+c^2-a^2}{2 b c}$ Here, $A=60^{\circ}$ $$ \begin{aligned} & \Rightarrow \cos 60^{\circ}=\frac{b^2+c^2-a^2}{2 b c} \Rightarrow \frac{1}{2}=\frac{b^2+c^2-a^2}{2 b c} \\ & \Rightarrow b c=b^2+c^2-a^2 \\ & \Rightarrow b^2+c^2=b c+a^2...(i) \end{aligned} $$ Now, $$ \begin{aligned} & \frac{b}{c+a}+\frac{c}{a+b}=\frac{b(a+b)+c(c+a)}{(c+a)(a+b)} \\ & =\frac{a b+b^2+c^2+a c}{a c+a^2+b c+a b} \\ & =\frac{a b+b c+a^2+a c}{a c+a^2+b c+a b} \quad[\because \text { from Eq. (i) }] \\ & =1 \end{aligned} $$

Hence, the correct answer is option: 'C'.

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