8,9 Let $M C$ be the perpendicular to the plane of triangle $A B C$. Is it true that $\angle A M B < \angle A C B$?
Consider an isosceles triangle $ABC$, where $AB = BC = 1$, $\angle ACB = \angle BAC = 1^{\circ}$, and $BK$ is the altitude. Then

$$
AC = 2AK = 2 \cos 1^{\circ} > 2 \cos 30^{\circ} = \sqrt{3}
$$

Let $MC = 1$. Then

$$
\angle BMC = 45^{\circ}, \angle AMC > 60^{\circ},
$$

Thus,

$$
\angle AMB + \angle BMC = \angle AMB + 45^{\circ} > \angle AMC > 60^{\circ}.
$$

Therefore,

$$
\angle AMB > 60^{\circ} - 45^{\circ} = 15^{\circ} > 1^{\circ} = \angle ACB.
$$

## Answer

No.