MHT CET · Maths · Properties of Triangles
In a \(\Delta \mathrm{ABC}\) if \(2 \cos \mathrm{C}=\operatorname{Sin} \mathrm{B} .\) CosecA, then
- A \(a=b\)
- B \(\mathrm{b}=\mathrm{c}\)
- C \(\mathrm{a}=\mathrm{c}\)
- D \(a=b=c\)
Answer & Solution
Correct Answer
(C) \(\mathrm{a}=\mathrm{c}\)
Step-by-step Solution
Detailed explanation
(D)
We know \(\frac{\sin B}{\sin A}=\frac{b}{a}\) and \(\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}\)
Given, \(2 \cos C=\sin B \cdot \operatorname{cosec} A \Rightarrow 2 \cos C=\frac{\sin B}{\sin A}\)
\(\therefore \frac{2\left(a^{2}+b^{2}-c^{2}\right)}{2 a b}=\frac{b}{a} \Rightarrow a^{2}+b^{2}-c^{2}=b^{2} \Rightarrow a^{2}=c^{2} \Rightarrow a=c\)
We know \(\frac{\sin B}{\sin A}=\frac{b}{a}\) and \(\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}\)
Given, \(2 \cos C=\sin B \cdot \operatorname{cosec} A \Rightarrow 2 \cos C=\frac{\sin B}{\sin A}\)
\(\therefore \frac{2\left(a^{2}+b^{2}-c^{2}\right)}{2 a b}=\frac{b}{a} \Rightarrow a^{2}+b^{2}-c^{2}=b^{2} \Rightarrow a^{2}=c^{2} \Rightarrow a=c\)
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