Let $S_{1}, S_{2}$ be planes perpendicular to each other, and let there be a half-line $e$ in their line of intersection, with the endpoint of $e$ being $O$. In one of the four quadrants determined by $S_{1}, S_{2}$, there are two half-lines starting from $O$, which form a $60^{\circ}$ angle with $e$, and the angles they form with $S_{1}$ are $30^{\circ}$ and $45^{\circ}$, respectively. What angle does the plane determined by the two half-lines form with the plane $S_{1}$?
I. Solution. Consider the cone with vertex $O$, axis $e$, and a half-aperture angle of $60^{\circ}$. According to the problem, both lines lie on the cone's surface. Let's choose the cone's generator to be of unit length, and let the line forming a $45^{\circ}$ angle with the plane $S_{1}$ be $O B$, and the line forming a $30^{\circ}$ angle with the plane $S_{1}$ be $O C$, where $O B$ and $O C$ are generators of the cone. Let the center of the base of the cone be $M$. Project the points $B$ and $C$ perpendicularly onto the plane $S_{1}$, and let the projections be $D$ and $E$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_ab60bbba7488993eb39fg-1.jpg?height=628&width=656&top_left_y=388&top_left_x=713)

Figure 1

Our task is to determine the angle between the planes of triangles $O B C$ and $O D E$. Since the triangle $O D E$ is the orthogonal projection of the triangle $O B C$, its area is the area of the triangle $O B C$ multiplied by the cosine of the angle between the two planes, denoted by $\varphi$:

$$
\cos \varphi=\frac{T_{O D E}}{T_{O B C}}
$$

Since the generator is of unit length, we can easily determine the lengths of the legs in the right triangles $O B M, O B D$, and $O C D$:

$$
O M=\cos 60^{\circ}=\frac{1}{2}, \quad B D=\sin 45^{\circ}=\frac{\sqrt{2}}{2}, \quad C E=\sin 30^{\circ}=\frac{1}{2}
$$

and

$$
O D=\cos 45^{\circ}=\frac{\sqrt{2}}{2}, \quad O E=\cos 30^{\circ}=\frac{\sqrt{3}}{2}
$$

We know two sides of the triangles $O B C$ and $O D E$, and we can calculate the third side from the above data. First, let's determine $D E$:

$$
\begin{gathered}
M E=\sqrt{O E^{2}-M O^{2}}=\frac{\sqrt{2}}{2}, \quad M D=\sqrt{O D^{2}-O M^{2}}=\frac{1}{2} \\
D E=M E-M D=\frac{\sqrt{2}-1}{2}
\end{gathered}
$$

Thus,

$$
T_{O D E}=\frac{1}{2} \cdot D E \cdot O M=\frac{1}{8}(\sqrt{2}-1)
$$

To determine $B C$, project $C$ perpendicularly onto $B D$. Let the projection be $F$.

$$
C F=E D=\frac{\sqrt{2}-1}{2}, \quad B F=B D-F D=B D-C E=\frac{\sqrt{2}-1}{2}=C F .
$$

We have found that $B F C$ is an isosceles right triangle, so

$$
B C=\sqrt{2} C F=\frac{2-\sqrt{2}}{2}
$$

The height of the isosceles triangle $O B C$ is $\sqrt{O B^{2}-\left(\frac{B C}{2}\right)^{2}}=\frac{\sqrt{10+4 \sqrt{2}}}{4}$, and thus

$$
T_{O B C}=\frac{1}{2} \cdot \frac{2-\sqrt{2}}{2} \frac{\sqrt{10+4 \sqrt{2}}}{4}=\frac{1}{8}(\sqrt{2}-1) \cdot \sqrt{5+2 \sqrt{2}}
$$

Finally, from (1):

$$
\cos \varphi=\frac{\frac{1}{8}(\sqrt{2}-1)}{\frac{1}{8}(\sqrt{2}-1) \sqrt{5+2 \sqrt{2}}}=\frac{1}{\sqrt{5+2 \sqrt{2}}}=0.357
$$

from which we get $\varphi=69^{\circ} 4^{\prime}$.

Remark. Let the intersection point of the lines $B C$ and $D E$ be $G$. The triangle $B D G$ is similar to the triangle $B F C$, and the latter is isosceles, so $D G=D B=\frac{\sqrt{2}}{2}$. Therefore, the triangle $O D G$ is also isosceles, and let the midpoint of its base be $H$. It is easy to see that the angle at $H$ in the triangle $B D H$ is equal to the desired angle $\varphi$. Since $O G^{2}=\frac{1}{4}+\frac{(\sqrt{2}+1)^{2}}{4}=\frac{2+\sqrt{2}}{2}$

$$
D H^{2}=O D^{2}-O H^{2}=\frac{1}{2}-\frac{2+\sqrt{2}}{8}=\frac{2-\sqrt{2}}{8}
$$

and thus

$$
\operatorname{tg} \varphi=\frac{2}{2-\sqrt{2}} \sqrt{4+2 \sqrt{2}}=2.613, \quad \varphi=69^{\circ} 4^{\prime}
$$

II. Solution. Let's set up a coordinate system such that the point $O$ is the origin of the coordinate system, the ray $e$ is the positive $X$-axis, and the plane of the $X$ and $Y$ axes is the plane $S_{1}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_ab60bbba7488993eb39fg-2.jpg?height=580&width=602&top_left_y=1109&top_left_x=748)

Figure 2

If the unit vector $e$ forms an angle $\alpha$ with the $X$-axis and an angle $\beta$ with the $X Y$ plane, then the components of the vector $e$ along the axes are, according to the diagram,

$$
\left(\cos \alpha, \varepsilon \sqrt{1-\cos ^{2} \alpha-\sin ^{2} \beta}, \sin \beta\right)
$$

Here, $\varepsilon$ must be chosen as +1 or -1 depending on which side of the plane $S_{2}$ the vector $e$ falls. According to this, if the direction vectors of the two rays in the problem are $\mathbf{a}$ and $\mathbf{b}$, then (2) gives

$$
\begin{aligned}
& \mathbf{a}=i \cos 60^{\circ}+j \varepsilon \sqrt{1-\cos ^{2} 60^{\circ}-\sin ^{2} 45^{\circ}}+k \cdot \sin 45^{\circ}=\frac{1}{2} i+\frac{\varepsilon}{2} j+\frac{\sqrt{2}}{2} k \\
& \mathbf{b}=i \cos 60^{\circ}+j \varepsilon \cdot \sqrt{1-\cos ^{2} 60^{\circ}-\sin ^{2} 30^{\circ}}+k \sin 30^{\circ}=\frac{1}{2} i+\varepsilon \frac{\sqrt{2}}{2} j+\frac{1}{2} k
\end{aligned}
$$

Here, $\varepsilon$ must be chosen the same for both, since the two rays fall into the same quadrant. The normal vector to the plane spanned by $\mathbf{a}$ and $\mathbf{b}$ is

$$
\mathbf{n}=\mathbf{a} \times \mathbf{b}=-\frac{\varepsilon}{4} i+\frac{\sqrt{2}-1}{4} j+\varepsilon \frac{\sqrt{2}-1}{4} k
$$

The normal vector to the plane $S_{1}$ is simply $k$. The angle between the plane and $S_{1}$ is the same as the angle between their normal vectors. The cosine of the angle between the normal vectors can be obtained from the definition of the dot product:

$$
|\mathbf{u} \cdot \mathbf{v}|=|\mathbf{u}| \cdot|\mathbf{v}| \cdot \cos (\mathbf{u}, \mathbf{v})-\text { so } \cos (\mathbf{u}, \mathbf{v})=\frac{|\mathbf{u} \cdot \mathbf{v}|}{|\mathbf{u}| \cdot|\mathbf{v}|}
$$

that is,

$$
\begin{aligned}
\cos \varphi & =\frac{\left|\left(-\frac{\varepsilon}{4} i+\frac{\sqrt{2}-1}{4} j+\varepsilon \frac{\sqrt{2}-1}{4} k\right) \cdot k\right|}{\left|-\frac{\varepsilon}{4} i+\frac{\sqrt{2}-1}{4} j+\varepsilon \frac{\sqrt{2}-1}{4} k\right| \cdot 1}= \\
& =\frac{|\varepsilon||\sqrt{2}-1|}{\sqrt{1^{2}+(\sqrt{2}-1)^{2}+(\sqrt{2}-1)^{2}}}=\frac{\sqrt{2}-1}{\sqrt{7-4 \sqrt{2}}}=0.357,
\end{aligned}
$$

from which $\varphi=69^{\circ} 4^{\prime}$.