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I have triangle $ABC$. How to determine the length of the sides $b$ and $c$? I know that $a=4.5cm$, $\alpha=2\beta$ and $b:c=3:7$.

$\frac{b}{c}=\frac{sin\beta}{sin\gamma}$

$\gamma=180-3\beta$

But still I don't have idea how to get values of sides?

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  • $\begingroup$ Is $\alpha$ the angle opposite side $a = 4.5 cm$? and similarly for $\beta,$ opposite the side with length $b$, and $\gamma$ opposite the side of length $c$? $\endgroup$ – Namaste Nov 15 '16 at 0:09
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$$\frac{b}{c} = \frac{sin \beta}{sin \gamma} = \frac{sin \beta}{sin(180-3\beta)}=\frac{1}{3-4sin^2\beta} = \frac{3}{7}$$ Solve this to get the value of $\beta$, hence $\gamma$ and $\alpha$. Then use sine rule for triangles to get the side lengths.

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$$b=3x,c=7x,a=4.5\\A=2B\\\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}\\cos B=\frac{a^2+c^2-b^2}{2ac}$$ $$\frac{sin2B}{4.5}=\frac{sinB}{3x}\\ \to \frac{2sinB.cosB}{4.5}=\frac{sinB}{3x} \\ \to 2cos B.3x=4.5 \\ cos B=\frac{4.5}{6x} $$ $$cos B=\frac{4.5^2+(7x)^2-(3x)^2}{2(4.5)(7x)}\\----\\ \to \\\frac{4.5}{6x}=\frac{4.5^2+(7x)^2-(3x)^2}{2(4.5)(7x)}$$now you can find $x$ $$\frac{2(4.5)^2(7x)}{6x}=\frac{4.5^2+(7x)^2-(3x)^2}{1}$$ so :then you have $b=3x,c=7x$

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