Question

Does anyone know the answer?

Answer 1

`m\angle \mathrm{AED}=(3 \pi)/(5)`

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:

`$\frac{\text { arc length }}{\text { circumference }}=\frac{\text { arc measure }}{\text { radians in a circle }}$`

`\Rightarrow (\pi)/(10 \pi)=\frac{\text { arc measure }}{2 \pi}`

⇒ arc measure =
`(\pi)/(5)`

To find the arc measure of CD:

`$\frac{\text { arc length }}{\text { circumference }}=\frac{\text { arc measure }}{\text { radians in a circle }}$`

`\Rightarrow (3\pi)/(10 \pi)=\frac{\text { arc measure }}{2 \pi}`

⇒ arc measure =
`(\pi)/(5)`)

⇒ arc measure =
`(3 \pi)/(5)`

The measure of an inscribed angle is half of the measure of the arc it intercepts.

arc measure of BA =
`(\pi)/(5)`, then
`\mathrm{m} \angle \mathrm{BCA}=(\pi)/(10)=0.1 \pi`

Similarly, arc measure of CD =
`(3 \pi)/(5)`, then
`\mathrm{m} \angle \mathrm{CBD}=(3 \pi)/(10)=0.3 \pi`

We know that sum of the interior angles of a triangle BCE = π

0.1π + 0.3π + m∠BEC = π

`\Rightarrow \mathrm{m} \angle \mathrm{BEC}=0.6 \pi`

`\mathrm{m} \angle \mathrm{BEC}=\mathrm{m} \angle \mathrm{AED}` (by vertical angle theorem)

Hence, the measure of
`\angle \mathrm{AED}=(3 \pi)/(5)`.