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In $\triangle ABC$, $A$ is the biggest angle, and $C$ is the smallest angle. Let $a$, $b$, $c$ be the length of the sides that are opposite from $\angle A$, $\angle B$, $\angle C$, respectively. If $\angle A=2\angle C$ and $a+c=2b$, find $a:b:c$.

I have no idea how to start, but Law of Sines/Law of Cosines could be useful.

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$$\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}$$

This implies that $$\frac{b}{\sin(B)}=\frac{a+c}{\sin(A)+\sin(C)}=\frac{2b}{\sin(A)+\sin(C)}$$

Combine this with $$A=2C\\ B=180^\circ -A -C =180^\circ -3C$$ and you can find $C$. From there it is easy

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  • $\begingroup$ How do you get from equation 1 to 2? $\endgroup$ – Baker013273213 Sep 26 at 23:36
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    $\begingroup$ @Baker013273213 If $\frac{a}{b}=\frac{c}{d}$ then they are also equal to $\frac{a+c}{b+d}$. This is sometimes called "Derived proportions". If you are not familiar, here is the proof: Let $x= \frac{a}{b}=\frac{c}{d}$. Then $$a=bx \\c=dx$$ and hence $$\frac{a+c}{b+d}=\frac{bx+dx}{b+d}=x$$ $\endgroup$ – N. S. Sep 27 at 2:17
  • $\begingroup$ Continuing on your progress, $\frac{1}{Sin{180-3C}}=\frac{2}{Sin {2C} +Sin {C}$ $\frac{1}{Sin{3C}}=\frac{2}{2 Sin {C} Cos {C} +Sin {C}$ How could we solve this? $\endgroup$ – Baker013273213 Sep 27 at 19:50

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