## [ The larger side is opposite the larger angle ]

 [Sum of the angles of a triangle. Theorem about the exterior angle.] In triangle $ABC$, it is known that $AB < BC < AC$, and one of the angles is twice as small as another and three times as small as the third. Find the angle at vertex $A$.
Against the larger side of a triangle lies the larger angle.

## Solution

Since in a triangle, the larger angle lies opposite the larger side, the smallest angle of triangle $A B C$ lies opposite side $A B$, that is, it is angle $A C B$. Let $\angle C=\gamma$. Then $\angle A=2 \gamma, \angle B=3 \gamma$. By the theorem of the sum of the angles of a triangle, $\gamma+2 \gamma+3 \gamma=180^{\circ}$, from which $\gamma=30^{\circ}$.

Therefore, $\angle A=2 \gamma=60^{\circ}$.

## Answer

$60^{\circ}$.