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The diagonal of an isosceles trapezoid is perpendicular to the lateral side. Find the acute angle and the larger base of the trapezoid if the smaller base is 3 and the height of the trapezoid is 2.

#
Let $AD$ be the larger base of the isosceles trapezoid $ABCD$, $CH$-- the height of the trapezoid, $CH=2$, $BC$-- the smaller base, $BC=3$, $AC \perp CD$. By the property of the height of a right triangle drawn from the vertex of the right angle,

$$
CH^2 = AH \cdot DH = \frac{1}{2}(AD + BC) \cdot \frac{1}{2}(AD - BC) = \frac{1}{4}(AD^2 - BC^2)
$$

or

$$
4 = \frac{1}{4}(AD^2 - 9)
$$

from which we find that $AD = 5$. Therefore,

$$
DH = \frac{1}{2}(AD - BC) = 1, \operatorname{tg} \angle ADC = \frac{CH}{DH} = 2
$$

## Answer

$5 ; \operatorname{arctg} 2$.