85. Point $O$ is the common vertex of two equal cones located on the same side of plane $\alpha$ such that only one generatrix of each cone ( $O A$ for one cone and $O B$ for the other) lies in plane $\alpha$. It is known that the angle between the heights of the cones is $\beta$, and the angle between the height and the generatrix of the cone is $\varphi$, with $2 \varphi < \beta$. Find the measure of the angle between the generatrix $O A$ and the base plane of the other cone, which contains point $B$.
85. The sought angle complements to $\pi / 2$ the angle between the line $O A$ and the axis of the second cone. Let $P$ and $Q$ be the centers of the bases of the given cones, and $S$ be the point where the planes of the bases of the cones intersect the perpendicular erected to the plane $O A B$ at point $O$ (Fig. 7). In the pyramid $S O A B$, it is known that $|O A|=|O B|$, $S O$ is perpendicular to the plane $O A B$, $O P$ and $O Q$ are perpendicular to $S B$ and $S A$ respectively, $\widehat{P O B}=\widehat{Q O A}=\varphi$, $\widehat{P O Q}=\beta$. We need to find $\widehat{P O A}$. Let $|O A|=|O B|=l,|A B|=a$. Then

$$
\begin{gathered}
|O P|=|O Q|=l \cos \varphi, \quad|S A|=|S B|=\frac{l}{\sin \varphi} \\
|S P|=|S Q|=|O P| \operatorname{ctg} \varphi=l \frac{\cos ^{2} \varphi}{\sin \varphi} \\
|P Q|=|A B| \frac{|S P|}{|S B|}=a \cos ^{2} \varphi
\end{gathered}
$$

On the other hand,

$$
|P Q|=2|O P| \sin \frac{\beta}{2}=2 l \cos \varphi \sin \frac{\beta}{2}
$$

From this,

$$
a \cos \varphi=2 l \sin \frac{\beta}{2}
$$

Now let's find $|P A|$:

$$
\begin{aligned}
& |P A|^{2}=|P B|^{2}+|A B|^{2}-2|P B| \cdot|A B| \cos \widehat{P B A}= \\
& \quad=l^{2} \sin ^{2} \varphi+a^{2}-2 l \sin \varphi \cdot a \frac{a \sin \varphi}{2 l}=l^{2} \sin ^{2} \varphi+a^{2} \cos ^{2} \varphi
\end{aligned}
$$

But if $\gamma=\widehat{P O A}$, then from $\triangle P O A$ we have:

$$
|P A|^{2}=l^{2} \cos ^{2} \varphi+l^{2}-2 l^{2} \cos \varphi \cos \gamma
$$

Equating the two expressions for $|P A|^{2}$ and considering (1), we find

$$
\begin{gathered}
\cos \gamma=\cos \varphi-\frac{2 \sin ^{2}-\frac{\beta}{2}}{\cos \varphi} \\
\text { Answer: } \frac{\pi}{2}-\arccos \left(\cos \varphi-\frac{2 \sin ^{2} \frac{\beta}{2}}{\cos \varphi}\right)
\end{gathered}
$$