Question

1. as shown in the figure below, $overline{ab}$ is the diameter of $odot o$. point $c$ is on $odot o$. if $mangle boc = 30^{circ}$, then $mangle c$ is ____. 6. as shown in the figure, $overline{ab}$ is the diameter of $odot o$, and $overline{cd}$ is the chord of $odot o$. $overrightarrow{ab}$ and $overrightarrow{cd}$ intersect at point $e$. if $ab = 2de$, and $mangle e = 18^{circ}$, then $mangle aoc=$____$^{circ}$.

Explanation:

Step1: Recall circle - angle relationship

In a circle, the measure of an inscribed - angle is half the measure of the central - angle subtended by the same arc. In \(\odot O\), \(\angle BOC\) is a central - angle and \(\angle C\) is an inscribed - angle subtended by arc \(BC\).

Step2: Apply the inscribed - angle theorem

The inscribed - angle theorem states that if \(\angle BOC\) is the central - angle and \(\angle C\) is the inscribed - angle subtended by the same arc \(BC\), then \(m\angle C=\frac{1}{2}m\angle BOC\).
Given \(m\angle BOC = 30^{\circ}\), then \(m\angle C=\frac{1}{2}\times30^{\circ}=15^{\circ}\).

for the second question:

Step1: Let \(r\) be the radius of the circle \(\odot O\), so \(AB = 2r\).

Since \(AB = 2DE\), then \(DE=r\). Let's connect \(OD\). In \(\triangle ODE\), \(OD = r\) and \(DE=r\), so \(\triangle ODE\) is an isosceles triangle.

Step2: Find the measure of \(\angle DOE\)

In isosceles triangle \(\triangle ODE\), given \(m\angle E=18^{\circ}\), and since \(OD = DE\), then \(m\angle DOE=m\angle E = 18^{\circ}\).

Step3: Use the exterior - angle property of a triangle

The exterior - angle of a triangle is equal to the sum of the two non - adjacent interior angles. In \(\triangle OCD\), \(\angle AOC\) is an exterior angle of \(\triangle ODE\) (where \(\angle AOC\) and \(\angle DOE\) are related to the angles formed by the chords and the center of the circle).
The measure of \(\angle AOC\) is the sum of the measures of \(\angle ODE\) and \(\angle E\). Since \(\triangle ODE\) is isosceles with \(OD = DE\), \(m\angle ODE=m\angle DOE + m\angle E\).
We know that \(m\angle DOE = 18^{\circ}\) and \(m\angle E=18^{\circ}\), so \(m\angle AOC = 36^{\circ}\).

Answer:

\(15^{\circ}\)