A ray of an angle is tangent to a given circle of radius $r$, and the other ray intersects the circle, cutting off segments of lengths $a$ and $b$. What is the measure of the angle?
I. Solution: Our task is to establish a relationship between $a$ and the given $a, b$, and $r$, from which we can calculate $\alpha$. This can be achieved as follows.

Draw the angle with vertex $A$ and the circle centered at $O$ (Figure 1), and project the farther intersection point $C$ of the secant ray from $A$ onto the tangent ray, and the center of the circle onto the projection ray $C F$ and the tangent.

![](https://cdn.mathpix.com/cropped/2024_05_02_d8982a78c2addae5476cg-1.jpg?height=477&width=645&top_left_y=347&top_left_x=726)

The tangent segment $A E$ is the geometric mean of the segments $a$ and $b$ of the secant drawn from $A$: $A E=\sqrt{a b}$.

Calculate the sides of the right triangle $A C F$:

$$
A F=b \cos \alpha, \quad C F=b \sin \alpha
$$

Based on these, the sides of the right triangle $O C D$ are:

$$
O C=r, \quad O D=b \cos \alpha-\sqrt{a b}, \quad C D=b \sin \alpha-r
$$

Apply the Pythagorean theorem to this triangle, and we immediately get a suitable relationship between $\alpha$ and the given data:

$$
(b \sin \alpha-r)^{2}+(b \cos \alpha-\sqrt{a b})^{2}=r^{2}
$$

Expanding and rearranging the left side:

$$
\begin{gathered}
b^{2} \sin ^{2} \alpha-2 b r \sin \alpha+r^{2}+b^{2} \cos ^{2} \alpha-2 b \sqrt{a b} \cos \alpha+a b=r^{2} \\
b^{2}+a b-2 b(r \sin \alpha+\sqrt{a b} \cos \alpha)=0
\end{gathered}
$$

(When combining the $b^{2}$ terms, we used the fact that $\sin ^{2} \alpha+\cos ^{2} \alpha=1$.)

We can simplify by dividing by $b$. Express $\cos \alpha$ in terms of $\sin \alpha$ and rearrange the equation:

$$
\sqrt{a b} \sqrt{1-\sin ^{2} \alpha}=\frac{a+b}{2}-r \sin \alpha
$$

Squaring both sides of this equation:

$$
\left(r^{2}+a b\right) \sin ^{2} \alpha-r(a+b) \sin \alpha+\frac{(a-b)^{2}}{4}=0
$$

From this,

$$
\begin{aligned}
(\sin \alpha)_{1,2}= & \frac{r(a+b) \pm \sqrt{r^{2}(a+b)^{2}-\left(r^{2}+a b\right)(a-b)^{2}}}{2\left(r^{2}+a b\right)}= \\
& =\frac{r(a+b) \pm \sqrt{4 a b r^{2}-a b(a-b)^{2}}}{2\left(r^{2}+a b\right)}
\end{aligned}
$$

From the obtained value, we generally get two values for $\alpha$.

It is easily verifiable that the problem can be solved in the same way, leading to the same result, even if the center of the circle is outside the angle rays or if the angle is obtuse.

Dániel Gábor (Bp. Vlll., Piarista g. III. o. t.)

II. Solution: Let the two angles formed by the line $O A$ connecting the vertex of the angle to the center of the circle with the angle rays be $\alpha_{1}$ and $\alpha_{2}$. Then the desired angle $\alpha$ is either $\alpha=\alpha_{2}-\alpha_{1}$ or $\alpha=\alpha_{2}+\alpha_{1}$, depending on whether $O$ is outside or between the angle rays.

![](https://cdn.mathpix.com/cropped/2024_05_02_d8982a78c2addae5476cg-2.jpg?height=477&width=656&top_left_y=125&top_left_x=737)

For the angle $\alpha_{2}$, from the right triangle $A O E$ (Figure 2),

$$
\operatorname{tg} \alpha_{2}=\frac{O E}{A E}=\frac{r}{\sqrt{a b}}
$$

from which $\alpha_{2}$ can be found.

The angle $\alpha_{1}$ can be determined from the right triangle $A O F$, where $F$ is the projection of $O$ onto the secant ray. One of the legs of the triangle is

$$
A F=a+\frac{b-a}{2}=\frac{b+a}{2}
$$

The hypotenuse $A O$ can be calculated from the right triangle $A O E$:

$$
A O=\sqrt{r^{2}+a b}
$$

Thus,

$$
\cos \alpha_{1}=\frac{a+b}{2 \sqrt{r^{2}+a b}}
$$

$\alpha_{1}$ can be found from this.

The desired angle is therefore

$$
\alpha=\operatorname{arctg} \frac{r}{\sqrt{a b}} \pm \arccos \frac{a+b}{\sqrt{r^{2}+a b}}
$$

This solution applies to both acute and obtuse angles.