An observer sees the center of a balloon at an elevation angle of $45^{\circ}$, while another observer sees it at an elevation angle of $22.5^{\circ}$. The first observer is to the south, and the second is to the northwest of the balloon's base point, with a distance of 1600 meters between them. How high is the balloon floating above the horizontal ground?
I. Solution. Let the balloon be $L$, its orthogonal projection on the horizontal ground (foot point) be $T$; the observers be $A$ and $B$; the elevation angle of the balloon from point $A$ be $\alpha=45^{\circ}$, from point $B$ be $\beta=22.5^{\circ}$, the distance between the two observers be $A B=d=1600 \mathrm{~m}$, and the angle between the southern direction of $T A$ and the northwestern direction of $T B$ be $\delta=135^{\circ}$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_2e8c68c7b704e6ab6a56g-1.jpg?height=420&width=586&top_left_y=305&top_left_x=756)

Figure 1

- The right triangle $A T L$ is isosceles because $\alpha=45^{\circ}$, so $A T$ is equal to the desired height $T L$, which we denote as $m$. Draw a line from $L$ in the triangle $B T L$ that forms a $45^{\circ}$ angle with $L T$. Since $B L T \angle=90^{\circ}-L B T \angle=67.5^{\circ} > 45^{\circ}$, this line intersects the segment $B T$ at a point $C$. $T C=T L=m$ and $B L C \angle=B L T \angle-45^{\circ}=22.5^{\circ}=T B L \angle=C B L \angle$, so the triangle $B C L$ is isosceles: $B C=C L=m \sqrt{2}$.

Let the orthogonal projection of point $A$ on the line $B T$ be $D$. Then the right triangle $A D T$ is also isosceles because the angle at $T$ in $A T D \angle=180^{\circ}-A T B \angle=45^{\circ}$, so $A D=D T=m / \sqrt{2}$.

The hypotenuse of the right triangle $A B D$ is given: $d=1600 \mathrm{~m}$, and the legs can be expressed in terms of the desired height $m$: $A D=m / \sqrt{2}, B D=B C+C T+T D=m \sqrt{2}+m+m / \sqrt{2}$. Thus, by the Pythagorean theorem,

$$
A D^{2}+D B^{2}=\left(\frac{m}{\sqrt{2}}\right)^{2}+\left(m \sqrt{2}+m+\frac{m}{\sqrt{2}}\right)^{2}=A B^{2}=1600^{2}
$$

From this,

$$
m^{2}\left(\frac{1}{2}+2+2 \sqrt{2}+1+2+\sqrt{2}+\frac{1}{2}\right)=m^{2} 3 \sqrt{2}(\sqrt{2}+1)=1600^{2}
$$

or

$$
m=\frac{1600}{\sqrt{3 \sqrt{2}(\sqrt{2}+1)}}=\frac{1600 \sqrt{\sqrt{2}-1}}{\sqrt{3 \sqrt{2}}}=\frac{800}{3} \sqrt{\sqrt{8}(\sqrt{2}-1) \cdot 3}=\frac{800}{3} \sqrt{12-\sqrt{72}} \approx 500 \mathrm{~m}
$$

(rounded to the nearest meter).

II. Solution. We solve the problem generally, for any position of observers $A$ and $B$ on the horizontal plane through the foot point and for any acute angles $\alpha$ and $\beta$.

From the right triangles $A T L$ and $B T L$, $A T=m \operatorname{ctg} \alpha, B T=m \operatorname{ctg} \beta$, and thus from the triangle $A T B$ by the cosine rule,

$$
\begin{aligned}
d^{2} & =m^{2} \operatorname{ctg}^{2} \alpha+m^{2} \operatorname{ctg}^{2} \beta-2 m^{2} \operatorname{ctg} \alpha \operatorname{ctg} \beta \cos \delta= \\
& =m^{2}\left(\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta-2 \operatorname{ctg} \alpha \operatorname{ctg} \beta \cos \delta\right)
\end{aligned}
$$

and from this,

$$
m=\frac{d}{\sqrt{\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta-2 \operatorname{ctg} \alpha \operatorname{ctg} \beta \cos \delta}}
$$

(If $A B T$ is a real triangle, then the expression under the square root is positive, as the cosine rule is valid.)

With the data $d=1600 \mathrm{~m}, \alpha=45^{\circ}, \beta=22.5^{\circ}, \delta=135^{\circ}$, $m \approx 500 \mathrm{~m}$.