12.087. Find the angle between the slant height and the height of a cone, the lateral surface of which is the mean proportional between the area of the base and the total surface area.
## Solution.

According to the problem, $S_{\text {bok }}=\sqrt{S_{\text {osn }} \cdot S}=\sqrt{S_{\text {osn }}\left(S_{\text {bok }}+S_{\text {osn }}\right)}$. Let the slant height of the cone be $l$, and the radius of the base circle be $R$. Then $S_{\text {bok }}=\pi R l, S_{\text {osn }}=\pi R^{2}$, and $\sin \alpha=\frac{R}{l}$, where $\alpha$ is the angle to be found, so $R=l \sin \alpha$. Therefore, $\left.\pi l^{2} \sin \alpha=\sqrt{\pi l^{2} \sin \alpha\left(\pi l^{2} \sin \alpha+\pi l^{2} \sin ^{2} \alpha\right.}\right)$, from which $\sin ^{2} \alpha+\sin \alpha-1=0$. Solving this quadratic equation, we get $\sin \alpha=\frac{-1 \pm \sqrt{5}}{2}$. Since $0<\alpha<\frac{\pi}{2}$, then $\sin \alpha=\frac{\sqrt{5}-1}{2}$. Thus, $\alpha=\arcsin \frac{\sqrt{5}-1}{2}$.

Answer: $\arcsin \frac{\sqrt{5}-1}{2}$.