1. The distance from a point lying in a horizontal plane to the base of a television tower is 100 m. From this point, the tower (from base to top) is visible at an angle of $46^{\circ}$. Without using tables, calculators, or other computational devices, determine which is greater: the height of the tower or 103.3 m?
Answer: The height of the tower is greater. Solution. The height of the tower is $H=100 \cdot \operatorname{tg} 46^{\circ}$

$=100 \cdot \operatorname{tg}\left(45^{\circ}+1^{\circ}\right)=100 \frac{\operatorname{tg} 45^{\circ}+\operatorname{tg} 1^{\circ}}{1-\operatorname{tg} 45^{\circ} \cdot \operatorname{tg} 1^{\circ}}=100 \frac{1+\operatorname{tg} \frac{\pi}{180}}{1-\operatorname{tg} \frac{\pi}{180}}$. Since for $x \in\left(0 ; \frac{\pi}{2}\right)$ the inequality $\operatorname{tg} x>x$ holds, and also since $\pi>3$, then $H>100 \frac{1+\frac{\pi}{180}}{1-\frac{\pi}{180}}=100 \frac{180+\pi}{180-\pi}$

$>100 \frac{180+3}{180-3}=\frac{6100}{59}>103.3$. A more accurate estimate could have been given (since $\pi>3.14$): $H>100 \frac{180+\pi}{180-\pi}>100 \frac{180+3.14}{180-3.14}>103.55$.

The solution provided above is absolutely rigorous and does not use approximate calculations. An approximate calculation of the height of the tower is also possible: since for small angles $\operatorname{tg} x \approx x$, then $H=100 \cdot \operatorname{tg} 46^{\circ}=100 \frac{1+\operatorname{tg} \frac{\pi}{180}}{1-\operatorname{tg} \frac{\pi}{180}} \approx 100 \frac{1+\frac{\pi}{180}}{1-\frac{\pi}{180}}=100 \frac{180+\pi}{180-\pi} \approx 100 \frac{180+3.14}{180-3.14} \approx 103.55 \mathrm{M}$. This is a fairly accurate result (calculation by tables gives $100 \cdot \operatorname{tg} 46^{\circ} \approx 103.5530$), but such a solution cannot be considered flawless if the accuracy of the calculations is not assessed.

Answer to variant 162: 193.5 m is greater.

Answer to variant 163: the height of the tower is greater.

Answer to variant 164: 96.8 m is greater.