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

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(C) Given the expression $E = \frac{\sin A \cot C + \cos A}{\sin B \cot C + \cos B}$.Substituting $\cot C = \frac{\cos C}{\sin C}$,we get $E = \frac{\

Vedclass · Accepted Answer

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

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(C) Given the expression $E = \frac{\sin A \cot C + \cos A}{\sin B \cot C + \cos B}$.Substituting $\cot C = \frac{\cos C}{\sin C}$,we get $E = \frac{\frac{\sin A \cos C + \cos A \sin C}{\sin C}}{\frac{\sin B \cos C + \cos B \sin C}{\sin C}} = \frac{\sin(A+C)}{\sin(B+C)}$.Since $A+B+C = \pi$,we have $A+C = \pi - B$ and $B+C = \pi - A$.Thus,$E = \frac{\sin(\pi - B)}{\sin(\pi - A)} = \frac{\sin B}{\sin A} = \frac{b}{a}$.Given $b^2 = ac$,the sides $a, b, c$ are in a Geometric Progression $(G.P.)$. Let $b = ar$ and $c = ar^2$.Then $E = \frac{b}{a} = \frac{ar}{a} = r$.For a triangle to exist,the sum of any two sides must be greater than the third side:$1) a+b > c$ $\Rightarrow a+ar > ar^2$ $\Rightarrow 1+r > r^2$ $\Rightarrow r^2-r-1  0$,$r \in \left(0, \frac{1+\sqrt{5}}{2}\right)$.$2) a+c > b$ $\Rightarrow a+ar^2 > ar$ $\Rightarrow r^2-r+1 > 0$. This is true for all $r$.$3) b+c > a$ $\Rightarrow ar+ar^2 > a$ $\Rightarrow r^2+r-1 > 0$. Solving this gives $r > \frac{-1+\sqrt{5}}{2}$ (since $r > 0$).Combining these,$r \in \left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$.

Suppose that the sides $a, b, c$ of a triangle $ABC$ satisfy $b^2 = ac$. 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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