The vertex $B$ of an isosceles $\triangle ABC$ $(AB=BC)$ does NOT lie on the plane $\alpha$, but the base $AC=18$ of the triangle lies in the plane. The angle between $CB$ and $\alpha$ is $60^\circ$. Find the angle $\measuredangle(BM,\alpha)$ where $BM$ is the median of the triangle and $AB=15$.
Is the given angle $\measuredangle ACB=60^\circ$? Because the orthogonal projection of $BC$ is the line $MC$ or not? Is the orthogonal projection of the point $B$ the point $M$ in $\alpha$ and why? I am not sure I understand which angle I am supposed to find. We can use the fact that $BM\perp AC$ because the triangle is isosceles, but what else? Thank you in advance!
Imagine isosceles $\triangle ABC$ as being vertical on the plane $\alpha$, $AC$ contained in $\alpha$. $\triangle BMC$ is a $3-4-5$ triangle right-angled at $M$.
Now tilt the triangle, keeping $AC$ fixed, so $B$ moves closer to the plane. Call the projection of $B$, $P$. As $ABC$ is isosceles, $P$ will move on the perpendicular bisector of $AC$, away from $M$. Clearly in triangle $BPC$, right-angled at $P$, $\angle BCP$ is the angle between $BC$ and $\alpha$.
Initially in the vertical position, $P$ was at $M$ and $\angle BCP \approx 53.1^{\circ}$. As $B$ moves closer to the plane, this angle will reduce and go to zero when $B$ finally lies in the plane.
Thus the data $\angle BCP= 60^{\circ}$ is false; it is not compatible with $BC=15, AC=18$. $\angle BCP\le \arcsin (4/5)\approx 53.1^{\circ}$ always.