Paxton Hoffman

Answered

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.

Answer & Explanation

dtal50

Expert

2022-07-24Added 10 answers

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, ${K}^{\prime}\equiv K$.

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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