Answer:
[tex]m\angle \mathrm{AED}=\frac{3 \pi}{5}[/tex]
Step-by-step explanation:
Given radius = 5 units, arc length of BA = π, arc length of CD = 3π
Circumference of a circle = 2πr
= 2π(5)
= 10π
To find the arc measure of BA:
[tex]$\frac{\text { arc length }}{\text { circumference }}=\frac{\text { arc measure }}{\text { radians in a circle }}$[/tex]
[tex]\Rightarrow \frac{\pi}{10 \pi}=\frac{\text { arc measure }}{2 \pi}[/tex]
⇒ arc measure = [tex]\frac{\pi}{5}[/tex]
To find the arc measure of CD:
[tex]$\frac{\text { arc length }}{\text { circumference }}=\frac{\text { arc measure }}{\text { radians in a circle }}$[/tex]
[tex]\Rightarrow \frac{3\pi}{10 \pi}=\frac{\text { arc measure }}{2 \pi}[/tex]
⇒ arc measure = [tex]\frac{\pi}{5}[/tex])
⇒ arc measure = [tex]\frac{3 \pi}{5}[/tex]
The measure of an inscribed angle is half of the measure of the arc it intercepts.
arc measure of BA = [tex]\frac{\pi}{5}[/tex], then [tex]\mathrm{m} \angle \mathrm{BCA}=\frac{\pi}{10}=0.1 \pi[/tex]
Similarly, arc measure of CD = [tex]\frac{3 \pi}{5}[/tex], then [tex]\mathrm{m} \angle \mathrm{CBD}=\frac{3 \pi}{10}=0.3 \pi[/tex]
We know that sum of the interior angles of a triangle BCE = π
0.1π + 0.3π + m∠BEC = π
[tex]\Rightarrow \mathrm{m} \angle \mathrm{BEC}=0.6 \pi[/tex]
[tex]\mathrm{m} \angle \mathrm{BEC}=\mathrm{m} \angle \mathrm{AED}[/tex] (by vertical angle theorem)
Hence, the measure of [tex]\angle \mathrm{AED}=\frac{3 \pi}{5}[/tex].
