# Find an angle created by lateral edge and the base of the Pyramid

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..

## 1 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, $$K'\equiv K$$.

• Could you explain why that's the case? If $BK\perp{SK}$ wouldn't that mean that SK is the height of the pyramid? Jul 15, 2020 at 2:22
• The fact that $\measuredangle{SBK} = \measuredangle{SAK}$ is the same as $BK\perp{SK}$... But I don't see the reason... Jul 15, 2020 at 2:35