Problem 2. The distances from three points lying in a horizontal plane to the base of a television tower are 800 m, 700 m, and 500 m, respectively. From each of these three points, the tower is visible (from base to top) at a certain angle, and the sum of these three angles is $90^{\circ}$. A) Find the height of the television tower (in meters). B) Round the answer to the nearest whole number of meters.
Solution. Let the given distances be denoted by $a, b$, and $c$, the corresponding angles by $\alpha, \beta$, and $\gamma$, and the height of the tower by $H$. Then $\operatorname{tg} \alpha=\frac{H}{a}, \operatorname{tg} \beta=\frac{H}{b}, \operatorname{tg} \gamma=\frac{H}{c}$. Since $\frac{H}{c}=\operatorname{tg} \gamma$ $=\operatorname{tg}\left(90^{\circ}-(\alpha+\beta)\right)=\frac{1}{\operatorname{tg}(\alpha+\beta)}=\frac{1-\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}{\operatorname{tg} \alpha+\operatorname{tg} \beta}=\frac{1-\frac{H}{a} \cdot \frac{H}{b}}{\frac{H}{a}+\frac{H}{b}}$, we have $\frac{H}{c}\left(\frac{H}{a}+\frac{H}{b}\right)=1-\frac{H}{a} \cdot \frac{H}{b} \Leftrightarrow$ $\frac{H^{2}(a+b)}{a b c}=\frac{a b-H^{2}}{a b} \Leftrightarrow H^{2}(a+b)=c\left(a b-H^{2}\right) \Leftrightarrow H^{2}(a+b+c)=a b c \Leftrightarrow$ $H=\sqrt{\frac{a b c}{a+b+c}}$. For the given numerical data: $H=\sqrt{\frac{800 \cdot 500 \cdot 700}{800+500+700}}$ $=\sqrt{\frac{800 \cdot 350}{2}}=\sqrt{140000}=100 \sqrt{14}$.

Since $374=\sqrt{139876}<\sqrt{140000}<\sqrt{140250.25}=374.5$, we have $\sqrt{140000} \approx 374$.

A strict justification for the approximate answer (of the type indicated above or another) is required. Note that a conclusion of the form “since $140000-374^{2}<375^{2}-140000$, then $\sqrt{140000} \approx 374$” is unjustified.

Answer: A) $100 \sqrt{14}$ m; B) 374 m.

Answer for variant 212: A) $25 \sqrt{35}$; B) 148.