Question

9 as shown in the figure, a, b, c and d are on ⊙o, bc is a diameter of ⊙o. if m∠acb = 56°, then m∠adc is ____. a. 17° b. 34° c. 56° d. 68° 10 as shown in the figure below, bd is the diameter of ⊙o, and points a and c are on ⊙o. ab = ad, ac intersects bd at g. if m∠cod = 126°, then m∠agb is ____. a. 99° b. 108° c. 110° d. 117°

Explanation:

Step1: Recall the inscribed - angle theorem for question 9

The measure of an inscribed angle is half the measure of the central angle subtended by the same arc. Angles $\angle ACB$ and $\angle ADC$ subtend the same arc $\overset{\frown}{AB}$. So, $m\angle ADC=m\angle ACB = 56^{\circ}$.

Step2: Find the measure of $\angle AOD$ and $\angle AOB$ for question 10

Since $\overset{\frown}{AB}=\overset{\frown}{AD}$, the central angles $\angle AOB=\angle AOD$. The sum of the central - angles around a point is $360^{\circ}$, and given $m\angle COD = 126^{\circ}$, then $m\angle AOB+m\angle AOD=360^{\circ}- 126^{\circ}=234^{\circ}$. So, $m\angle AOB = m\angle AOD = 117^{\circ}$.

Step3: Find the measure of $\angle OAG$

The radius of the circle $OA = OD$, and the central angle $\angle AOD = 117^{\circ}$, so the inscribed angle $\angle OAG=\frac{1}{2}\angle AOD$. In $\triangle AOG$, we know that $\angle AOB = 117^{\circ}$, and the radius $OA = OB$. The measure of $\angle OAB=\angle OBA$. The measure of $\angle AGB$ is an exterior angle of $\triangle AOG$.
The measure of $\angle OAB=\frac{1}{2}(180 - 117)^{\circ}=31.5^{\circ}$.
The measure of $\angle AGB = 180-(180 - 117)\div2=108^{\circ}$.

Answer:

1. C. $56^{\circ}$
2. B. $108^{\circ}$