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There's a triangle $XYZ$, where $X, Y, Z$ are the angles and $x, y, z$ are the sides opposite to the angles respectively.

Given that $2X = 3Y$ and $x = 2z$, then find the possible value of $∠Z$.

I used the sine law, $$\frac{\sin X}{x} = \frac{\sin Y}{y} = \frac{\sin Z}{z}$$ and I got the following relation: $$\frac{\sin X}{\sin Z} = 2 = \frac{\sin(2X/3)}{y}$$

But I'm unable to solve further and get the value of $Z$. Please help.

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$\sin Z=\sin(\pi-X-Y)=\sin(X+Y)=\sin(5X/3)$, and $z=x/2$, so from $(\sin X)/x=(\sin Z)/z$ we get $(\sin X)/x=(\sin(5X/3))/(x/2)$ which is $\sin X=2\sin(5X/3)$. Can you take it from there?

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I'm not sure if I'm correct but I solved $sinX = 2sin(5X/3)$ which gave me $X = 6nπ$, now $sin(6nπ) = 0$, so $Z = 0$ which is not possible. Where am I going wrong sir? – Hyperbola Jul 18 '12 at 5:46
When $X$ is small, the left side is less than the right. When $X=\pi$, the left side is bigger than the right. So there's a solution somewhere in between. You can just try some angles for which it's easy to calculate both sines. – Gerry Myerson Jul 18 '12 at 7:08

Put $X:=3\alpha$, $Y:=2\alpha$. Then $Z=\pi-5\alpha$, and the sine law gives $$\sin(3\alpha)=2\sin(5\alpha)\ .$$ Writing $$\sin(3\alpha)=\sin(4\alpha)\cos\alpha-\cos(4\alpha)\sin\alpha\ ,\quad \sin(5\alpha)=\sin(4\alpha)\cos\alpha+\cos(4\alpha)\sin\alpha$$ we get the equation $$\tan(4\alpha)=-3\tan\alpha\ .\qquad(*)$$ Now $$\tan{\pi\over6}={1\over\sqrt{3}}\ , \qquad\tan(4{\pi\over6})=\tan({2\pi\over3})=-\tan({\pi\over3})=-\sqrt{3}\ .$$ Therefore $\alpha={\pi\over6}$ solves $(*)$, which makes $Z={\pi\over6}$. Thus the considered triangle is half of an equilateral triangle.

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