In the equilateral triangle $ABC$, the sides $BC$, $CA$, and $AB$ are seen from a point $P$ in the plane of the triangle at angles of $30^{\circ}$, $50^{\circ}$, and $80^{\circ}$, respectively. Calculate the distances $PA$, $PB$, and $PC$ if $AB=1 \text{~km}$.
Those points from which the side $BC$ is seen at an angle of $30^{\circ}$ lie on the arcs of the circles passing through $B$ and $C$ and corresponding to a central angle of $2 \cdot 30^{\circ} = 60^{\circ}$. Therefore, the centers of these two arcs, together with the points $B$ and $C$, form the vertices of two equilateral triangles. Thus, the center of one arc is $A$, and the center of the other is the reflection of $A$ across $BC$, denoted as $A'$. The arc $k$ of the first circle lies on the side of the line $BC$ that contains $A$, and the arc $k'$ of the second circle lies on the side of the line $BC$ that does not contain $A$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_237a236e5c93ec314b27g-1.jpg?height=367&width=563&top_left_y=394&top_left_x=770)

Figure 1

We will show that the angle subtended by the segment $AB$ from any point on $k'$ is less than $60^{\circ}$, so $P$ can only be on $k$, in which case $AP = AB = 1 \text{ km}$.

The points from which $AB$ is seen at an angle of $60^{\circ}$ lie on the larger arc of the circumcircle $k_1$ of triangle $ABC$ and on the arc of the reflection of $k_1$ across $AB$, denoted as $k_1'$. The points in the plane not covered by $k_1$ and $k_1'$ see the segment $AB$ at an angle less than $60^{\circ}$. Since the lines $A'B$ and $A'C$ are tangent to $k_1$, the rays of angle $B A' C$ separate the points of the arcs $k_1$ and $k_1'$, so the points of $k'$ are external to $k_1$; and the line $BC$ separates the points of $k'$ from $k_1'$. This proves our statement.

Since $AP = AB = AC$, the triangles $ABP$ and $ACP$ are isosceles with angles of $20^{\circ}$ and $80^{\circ}$ at $A$. The ray that forms a $20^{\circ}$ angle with $AB$ either passes inside triangle $ABC$ or outside. In the first case, it cannot form an $80^{\circ}$ angle with $AC$, so only the second case provides a solution. According to this, $P$ is obtained by measuring a $20^{\circ}$ angle from the ray $AB$ at $A$ on the side that does not contain $C$, and then measuring the segment $AP = AB$ on this ray. According to the above, the point thus obtained satisfies the requirements of the problem.

![](https://cdn.mathpix.com/cropped/2024_05_02_237a236e5c93ec314b27g-1.jpg?height=445&width=374&top_left_y=1393&top_left_x=862)

Figure 2

For the remaining segments $PB$ and $PC$, from the isosceles triangles $ABP$ and $ACP$ we have

$$
BP = 2AB \cos 80^{\circ} = 0.3472 \text{ km}, \quad CP = 1.2856 \text{ km}
$$