Question

given circle o with center d and diameter ac in the figure below, find each angle measure. find each angle measure. m(∠1) = □° m(∠2) = □° m(∠3) = □° m(∠4) = □° m(∠5) = □° (type whole numbers.)

Explanation:

Step1: Use property of isosceles triangle

Since \(DA = DB\) (radii of the same circle), \(\triangle DAB\) is isosceles. So \(m(\angle1)=m(\angle5) = 30^{\circ}\).

Step2: Find \(\angle ADB\)

In \(\triangle DAB\), using the angle - sum property of a triangle (\(180^{\circ}\) in a triangle), \(m(\angle ADB)=180^{\circ}- 30^{\circ}-30^{\circ}=120^{\circ}\). And \(m(\angle2)=180^{\circ}-m(\angle ADB)=60^{\circ}\).

Step3: Use inscribed - angle theorem

The inscribed - angle theorem states that an inscribed angle is half of the central angle subtended by the same arc. Arc \(AB\) subtends \(\angle ADB\) at the center and \(\angle3\) at the circumference. So \(m(\angle3)=\frac{1}{2}m(\angle ADB) = 30^{\circ}\).

Step4: Use angle in a semi - circle property

\(\angle ABC = 90^{\circ}\) (angle inscribed in a semi - circle). Since \(m(\angle ABC)=m(\angle4)+m(\angle5)\) and \(m(\angle5) = 30^{\circ}\), then \(m(\angle4)=90^{\circ}-30^{\circ}=60^{\circ}\).

Answer:

\(m(\angle1)=30^{\circ}\)
\(m(\angle2)=60^{\circ}\)
\(m(\angle3)=30^{\circ}\)
\(m(\angle4)=60^{\circ}\)
\(m(\angle5)=30^{\circ}\)