In a scalene triangle, one angle is twice another. If one of the angles is $36^{\circ}$, determine all possibilities for the angles of the triangle.

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Solution

Let $xx+y+z \\
& =180^{\circ} \\
z & >\frac{180^{\circ}}{3} \\
& =60^{\circ}
\end{aligned}
$$

Similarly, we can conclude that $xy$. If $z=2 x$, then $y=180^{\circ}-3 x=72^{\circ}$, and this is also a contradiction, as we again have $z=y$. Therefore, if $x=36^{\circ}$, then $z=2 y$ and $180^{\circ}=x+y+z=36^{\circ}+3 y$, that is, $(x, y, z)=\left(36^{\circ}, 48^{\circ}, 96^{\circ}\right)$. In the second case, when $y=36^{\circ}$, if $z=2 y$, we will have $x=180^{\circ}-3 y=72^{\circ}$. This is impossible, as $x<y$. If $z=2 x$, then $3 x=180^{\circ}-y=144^{\circ}$, that is, $x=48^{\circ}$. Again, a contradiction due to $x<y$. The only remaining possibility is $y=2 x$ and thus $z=180^{\circ}-y-y / 2=126^{\circ}$, that is, $(x, y, z)=\left(18^{\circ}, 36^{\circ}, 126^{\circ}\right)$.