Suppose that the sides a,b, c of a triangle A | vedclass

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Question

(c)

$\frac{\sin A \frac{\cos C}{\sin C}+\cos A}{\sin B \frac{\cos C}{\sin C}+\cos B}=\frac{\frac{\sin A \cos C+\sin C \cos A}{\sin C}}{\frac{\sin B

Vedclass · Accepted Answer

$\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$

Vedclass · Answer

(c) $\frac{\sin A \frac{\cos C}{\sin C}+\cos A}{\sin B \frac{\cos C}{\sin C}+\cos B}=\frac{\frac{\sin A \cos C+\sin C \cos A}{\sin C}}{\frac{\sin B \operatorname{Cos} C+\cos B \sin C}{\sin C}}$ $=\frac{\sin (A+C)}{\sin (B+C)}=\frac{b}{a}$ a, b, c $\rightarrow$ G.P. $b =a r ; c =a r ^2$ $C-I$ If $r > 1$ then $a+a r > a r^2 \Rightarrow r^2-r-1 < 0$ $r \in\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$ but $r > 1$ Hence $r \in\left(1, \frac{1+\sqrt{5}}{2}\right) \quad \ldots .( I ) \ldots$ $C-II$ $0 < r < 1$ $a r+a r^2 > a \Rightarrow r^2+r-1 > 0$ $r \in\left(-\infty, \frac{-1-\sqrt{5}}{2}\right) \cup\left(\frac{-1+\sqrt{5}}{2}, \infty\right)$ but $r \in(0,1)$ $r \in\left(\frac{\sqrt{5}-1}{2}, 1\right) \quad \ldots (II) ....$ $C-III$ when $r=1$ then $\Delta$ is equilateral Hence $r \in\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$

Suppose that the sides $a,b, c$ of a triangle $A B C$ satisfy $b^2=a c$. Then the set of all possible values of $\frac{\sin A \cot C+\cos A}{\sin B \cot C+\cos B}$ is

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