Paxton Hoffman

2022-07-23

Pyramid SABC has right triangular base ABC, with $\mathrm{\angle }ABC={90}^{\circ }$. Sides $AB=\sqrt{3},BC=3$. Lateral lengths are equal and are equal to 2. Find the angle created by lateral length and the base.

dtal50

Expert

Step 1
Let K be a middle point of AC.
Just $\measuredangle SBK=\measuredangle SAK=\mathrm{arccos}\frac{\sqrt{3}}{2}={30}^{\circ }.$.
BK is a median of $\mathrm{\Delta }ABC$ 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.
Step 2
Let SK′ be an altitude of the pyramid.
Thus, since $SA=SB=SC$, we obtain: ΔSAK′≅ΔSBK′ and $\mathrm{\Delta }SA{K}^{\prime }\cong \mathrm{\Delta }SC{K}^{\prime }$, which gives $A{K}^{\prime }=B{K}^{\prime }=C{K}^{\prime },$, which says K′ is a center of the circumcircle for $\mathrm{\Delta }ABC$.
Thus, ${K}^{\prime }\equiv K$.

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