Pyramid $SABC$ has right triangular base $ABC$, with $\angle{ABC}=90^\circ$. Sides $AB = \sqrt3, BC = 3$. Lateral lengths are equal and are equal to $2$. Find the angle created by lateral length and the base.
Here's my attempt, but I didn't get very far:
First we calculate the hypotenuse $AC = \sqrt{9+3}= 2\sqrt3$. The angle between the lateral length and the base will be the angle created by the slant height (or apothem) and the line towards it. If we draw a perpendicular from $SK$ to hypotenuse $AC$, the height will split the base in two, since $SAC$ is an isosceles triangle. After that we draw the line from point $B$ to $K$, the angle we're looking for will be $\angle{SBK}$.
I'm not sure how to continue after this, the thing is, I can probably calculate all 3 sides and use the cosine theorem from there, but the solution I saw to this problem said that line $BK$ will create a perpendicular with SK and from there on calculating the angle is trivial, but i'm not seeing how that's the case..
Best Answer
Let $K$ be a middle point of $AC$.
Just $$\measuredangle SBK=\measuredangle SAK=\arccos\frac{\sqrt3}{2}=30^{\circ}.$$ BK is a median of $\Delta ABC$ and is not perpendicular to $AC$, otherwise $AB=BC$, which is a contradiction.
By the way, $BK\perp SK$, but we said about it in the first line.
Let $SK'$ be an altitude of the pyramid.
Thus, since $SA=SB=SC$, we obtain: $\Delta SAK'\cong\Delta SBK'$ and $\Delta SAK'\cong\Delta SCK'$, which gives $$AK'=BK'=CK',$$ which says $K'$ is a center of the circumcircle for $\Delta ABC$.
Thus, $K'\equiv K$.