The height of a certain tower $C C_{1}$ is observed from the endpoints $A B=c$ of a horizontal distance at the same level as the footpoint $C$, with angles of elevation $\delta=12^{\circ} 05^{\prime}$ and $\varepsilon=6^{\circ} 55^{\prime}$, respectively. What is the height of $C C_{1}$ if $c=333.4 \, \text{m}$ and $\angle BAC=\alpha=103^{\circ} 21^{\prime}$?
The figure shows the layout.

![](https://cdn.mathpix.com/cropped/2024_05_02_d2b4b7291ecf9f2b9c69g-1.jpg?height=388&width=550&top_left_y=191&top_left_x=777)

The data provided is insufficient to directly solve either the right-angled triangles $A C C_{1}$ or $B C C_{1}$, or the triangle $A B C$. However, if we express the common leg $C C_{1}$ from both right-angled triangles, we can derive a relationship between $a$ and $b$, which will help us solve the triangle $A B C$.

From the right-angled triangles,

$$
C C_{1}=x=b \operatorname{tg} \delta=a \operatorname{tg} \varepsilon, \quad \text { from which } \quad \frac{a}{b}=\frac{\operatorname{tg} \delta}{\operatorname{tg} \varepsilon}
$$

Applying the sine rule to triangle $A B C$:

$$
\frac{a}{b}=\frac{\operatorname{tg} \delta}{\operatorname{tg} \varepsilon}=\frac{\sin \alpha}{\sin \beta}
$$

from which

$$
\sin \beta=\frac{\sin \alpha \operatorname{tg} \varepsilon}{\operatorname{tg} \delta}=\frac{\sin 103^{\circ} 21^{\prime} \operatorname{tg} 6^{\circ} 55^{\prime}}{\operatorname{tg} 12^{\circ} 05^{\prime}}
$$

Calculating the value of $\beta$ using logarithmic tables,

$$
\beta=33^{\circ} 17^{\prime} 30^{\prime \prime}
$$

and thus

$$
\gamma=43^{\circ} 11^{\prime} 30^{\prime \prime}
$$

From triangle $A B C$ using the sine rule,

$$
b=\frac{c \sin \beta}{\sin \gamma}
$$

From triangle $A C C_{1}$,

$$
\begin{aligned}
& x=b \operatorname{tg} \delta=\frac{c \sin \beta \operatorname{tg} \delta}{\sin \gamma}=\frac{333.4 \sin 33^{\circ} 27^{\prime} 30^{\prime \prime} \operatorname{tg} 12^{\circ} 05^{\prime}}{\sin 43^{\circ} 11^{\prime} 30^{\prime \prime}} \\
& \lg 334.4=2.5229 \\
& \lg \sin 33^{\circ} 27^{\prime} 30^{\prime \prime}=9.7414-10 \\
& \begin{array}{rll}
5 & \underline{1.3 \cdot 4} & 07 \\
&
\end{array} \\
& \begin{aligned}
\lg \operatorname{tg} 12^{\circ} 05^{\prime} & =9.3306-10 \\
\hline \sum^{\prime} & =11.5949-10 \\
-\lg \sin 43^{\circ} 11^{\prime} 30^{\prime \prime} & =9.8354 \mp 10 \\
\hline \lg x & =1.7595 \\
x & =57.48 \approx 57.5 \text{ m}
\end{aligned} \\
& \begin{array}{rll}
\frac{5}{29} & \frac{1.3 \cdot 4}{5.2} & \frac{7}{14} \frac{12}{6} \cdot 3.5=7 \\
275 & 0- &
\end{array} \\
& \frac{315}{306} \quad \frac{37}{6} \cdot 5 \approx 31
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_d2b4b7291ecf9f2b9c69g-1.jpg?height=139&width=539&top_left_y=1928&top_left_x=1064)

Pál Pejtsik (Bp. IX., Fáy g. III. o. t.)