# Condition to Have a 60 Degrees Angle

### Proof

We use the following lemma which is easily proved:

If $P+Q+R=0\;$ then

$\displaystyle\sin P+\sin Q+\sin R=-4\sin\frac{P}{2}\sin\frac{Q}{2}\sin\frac{R}{2}.$

Since $\sqrt{3}=\displaystyle\frac{\sin 60^{\circ}}{\cos 60^{\circ}},$ the given relation may be written as

$\displaystyle\sum_{cyc}(\sin A\cos 60^{\circ}-\cos A\sin 60^{\circ})=\sum_{cyc}\sin (A-60^{\circ})=0.$

By Lemma, $\displaystyle\prod_{cyc}\sin\frac{A-60^{\circ}}{2}=0,\;$ and thus at least one of the factors is $0.$

### Acknowledgment

This problem has been proposed by W. J. Blundon [1966, 1122, Am Math Monthly]. The solution is by A. Krishnamurthi [1968, E 1936, Am Math Monthly, p 405].
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