1. In a certain triangle, the sum of the tangents of the angles turned out to be 2016. Estimate (at least to the nearest degree) the magnitude of the largest of its angles.
Solution. One of the tangents must exceed 600. This is only possible for an angle very close to $90^{\circ}$. We will prove that it exceeds $89.5^{\circ}$. This is equivalent to the statement that $\operatorname{tg} 0.5^{\circ}>$ $1 / 600$.

Let's start with the equality $\sin 30^{\circ}=1 / 2$. Note that $\sin 2 x=2 \sin x \cos x$, so $\sin x=\frac{\sin 2 x}{2 \cos x}>\frac{\sin 2 x}{2}$ for acute angles. Therefore:

$\sin 32^{\circ}>1 / 2 ; \quad \sin 16^{\circ}>1 / 4 ; \quad \sin 8^{\circ}>1 / 8$

$\sin 4^{\circ}>1 / 16 ; \quad \sin 2^{\circ}>1 / 32 ; \quad \sin 1^{\circ}>1 / 64$

$\sin 0.5^{\circ}>1 / 128>1 / 600 ; \quad \operatorname{tg} 0.5^{\circ}>1 / 600$.

Thus, one of the angles of the triangle is in the interval from 89.5 to 90 degrees. Note that it may not be the largest; but in this case, the largest angle is less than $90.5^{\circ}$. Therefore, to the nearest degree, the largest angle is $90^{\circ}$ in any case.