The lateral edge of a regular triangular pyramid is $\sqrt{5}$, and the height of the pyramid is 1. Find the dihedral angle at the base.

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Let $M$ be the center of the base $ABC$ of a regular triangular pyramid $ABCD$, and $K$ be the midpoint of $BC$. Then $DM$ is the height of the pyramid, $AK$ is the height of the equilateral triangle $ABC$, and $AK$ passes through point $M$. Therefore, $MKD$ is the linear angle of the dihedral angle at the base of the pyramid. From the right triangle $ADM$, we find that

$$
AM=\sqrt{AD^2-DM^2}=\sqrt{5-1}=2
$$

Then $MK=\frac{1}{2} AM=1$. Thus, $MKD$ is an isosceles right triangle. Therefore, $\angle MKD=$ $45^{\circ}$.

## Answer

$45^{\circ}$.