In a Delta ABC,if 2 cos C = sin B cdot operat | vedclass

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Question

(C) Given the equation: $2 \cos C = \sin B \cdot \operatorname{cosec} A$Since $\operatorname{cosec} A = \frac{1}{\sin A}$,we have $2 \cos C = \frac{\s

Vedclass · Accepted Answer

$a = c$

Vedclass · Answer

(C) Given the equation: $2 \cos C = \sin B \cdot \operatorname{cosec} A$Since $\operatorname{cosec} A = \frac{1}{\sin A}$,we have $2 \cos C = \frac{\sin B}{\sin A}$.Using the Sine Rule,$\frac{\sin B}{\sin A} = \frac{b}{a}$.Thus,$2 \cos C = \frac{b}{a}$.Using the Cosine Rule,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.Substituting this into the equation: $2 \left( \frac{a^2 + b^2 - c^2}{2ab} \right) = \frac{b}{a}$.$\frac{a^2 + b^2 - c^2}{ab} = \frac{b}{a}$.Multiplying both sides by $ab$,we get $a^2 + b^2 - c^2 = b^2$.$a^2 - c^2 = 0 \Rightarrow a^2 = c^2$.Since $a$ and $c$ are side lengths,$a = c$.

In a $\Delta ABC$,if $2 \cos C = \sin B \cdot \operatorname{cosec} A$,then:

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