October
18, 2006
The sides of a triangle have lengths 4, 5, and 6. Show that one of its
angles is twice another.
Solution. Let $a=4$,
$b=5$, and $c=6$. So then $C$ is the largest and $A$ the smallest angle
of the triangle. Then the Law of Cosines gives
$\begin{eqnarray}
c^2=a^2 &+& b^2-2ab\cos C \\
\cos C = \frac{a^2+b^2-c^2}{2ab} &=&
\frac{4^2+5^2-6^2}{2(4)(5)} =
\frac5{40} = \frac18
\end{eqnarray}$
and similarly, $\cos A = \textstyle\frac34$. Then $\cos 2A =
2\cos^2 A - 1 = 2(3/4)^2-1 = 2(9/16)-1 = 18/16-1 = 1/8$. Since
both $A$ and $C$ are between 0 and $\pi$, this forces $2A=C$.