In a $\Delta \mathrm{ABC}$ if $2 \cos \mathrm{C}=\operatorname{Sin} \mathrm{B} .$ CosecA, then

In a $\Delta \mathrm{ABC}$ if $2 \cos \mathrm{C}=\operatorname{Sin} \mathrm{B} .$ CosecA, then
  1. $a=b$
  2. $\mathrm{b}=\mathrm{c}$
  3. $\mathrm{a}=\mathrm{c}$
  4. $a=b=c$

Solution

We know $\frac{\sin B}{\sin A}=\frac{b}{a}$ and $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$ Given, $2 \cos C=\sin B \cdot \operatorname{cosec} A \Rightarrow 2 \cos C=\frac{\sin B}{\sin A}$ $\therefore \frac{2\left(a^{2}+b^{2}-c^{2}\right)}{2 a b}=\frac{b}{a} \Rightarrow a^{2}+b^{2}-c^{2}=b^{2} \Rightarrow a^{2}=c^{2} \Rightarrow a=c$

Asked in: MHT CET 2020 (16 Oct Shift 2)

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