8. In a circle, three chords $A A_{1}, B B_{1}, C C_{1}$ intersect at one point. The angular measures of the arcs $A C_{1}, A B, C A_{1}$, and $A_{1} B_{1}$ are $150^{\circ}, 30^{\circ}, 60^{\circ}$, and $30^{\circ}$, respectively. Find the angular measure of the arc $B_{1} C_{1}$.
Solution: Let's formulate several auxiliary statements.

1) Let the angular measure of the arc $AB$ (Fig.1) be $\varphi$. (This means that $\varphi$ is equal to the corresponding central angle $AOB$.) Then the length of the chord

![](https://cdn.mathpix.com/cropped/2024_05_06_09f4d2a1d6569731a9d0g-2.jpg?height=311&width=323&top_left_y=907&top_left_x=1569)
$AB=2R\sin(\varphi/2)$. Here $R$ is the radius of the circle.

2) Let two chords $AA_1$ and $BB_1$ intersect at point $T$ (Fig.2). The angular measures of the arcs $AB$ and $A_1B_1$ are $\varphi$ and $v$. Triangles $ATB$ and $A_1TB_1$ are similar by two angles (equal angles are marked). The similarity ratio

$$
k=AB / A_1B_1=\sin(\varphi/2) / \sin(v/2)
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_09f4d2a1d6569731a9d0g-2.jpg?height=254&width=220&top_left_y=1643&top_left_x=193)

Fig. 1

![](https://cdn.mathpix.com/cropped/2024_05_06_09f4d2a1d6569731a9d0g-2.jpg?height=411&width=411&top_left_y=1551&top_left_x=480)

Fig. 2

![](https://cdn.mathpix.com/cropped/2024_05_06_09f4d2a1d6569731a9d0g-2.jpg?height=480&width=488&top_left_y=1505&top_left_x=955)

Fig. 3

3) Referring to Fig. 3. Three chords intersect at one point, denoted as $T$. The angular measures of the six resulting arcs are marked on the figure. From the similarity of triangles $ATB$ and $A_1TB_1$ it follows (see point 2) the equality $AT / B_1T = \sin(\varphi/2) / \sin(v/2)$. Similarly, $\triangle B T C \sim \Delta B_1 T C_1 \Rightarrow B_1 T / C T = \sin(\psi/2) / \sin(u/2), \Delta C T A_1 \sim \Delta A T C_1 \Rightarrow C T / A T = \sin(\theta/2) / \sin(w/2)$. Multiplying the last three equalities, we get:

$1 = AT / B_1T \cdot B_1T / CT \cdot CT / AT = \sin(\varphi/2) / \sin(v/2) \cdot \sin(\psi/2) / \sin(u/2) \cdot \sin(\theta/2) / \sin(w/2)$.

Thus, the necessary (and in fact sufficient) condition for three chords to intersect at one point is the equality:

$$
\sin(\varphi/2) \sin(\theta/2) \sin(\psi/2) = \sin(u/2) \sin(v/2) \sin(w/2)
$$

Now it is not difficult to obtain the answer to the problem. Substituting the given data into this relation $w=150^\circ, \varphi=30^\circ, \theta=60^\circ, v=30^\circ$, and expressing $u$ from the equality $\varphi + u + \theta + v + \psi + w = 360^\circ$, we get the equation for determining the desired angle $\psi$:

$$
\sin 15^\circ \sin(\psi/2) \sin 30^\circ = \sin 15^\circ \sin \left(\left(90^\circ - \psi\right)/2\right) \sin 75^\circ
$$

From this, it is not difficult to obtain that $\psi = 60^\circ$.

Answer: $60^\circ$.