Question

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Find each indicated measure for circle O

Answer 1

Answer:

Angle A = 41, Arc CE = 42, Angle C = 41, Angle D = 40, Angle ABE = 61

Step-by-step explanation:

Angle A = Arc BD/2  (Inscribed Angle)

Angle A = 82/2

Angle A = 41

Arc CE = 2 x Angle CEB (Inscribed Angle)

Arc CE = 2 x 21

Arc CE = 42

Angle C = Arc BD/2 (Inscribed Angle)

Angle C = 82/2

Angle C = 41

Angle D = Arc AC/2  (Inscribed Angle)

Angle D = 80/2

Angle D = 40

Angle ABE = Arc AE/2

Angle ABE = Arc AC + Arc CE/2

Angle ABE = 122/2

Angle ABE = 61

Answer 2

m < A =  $39^{0}$ , mCE =  $40^{0}$, m < c = $39^{0}$, m < D = $37^{0}$ m < ABE = $57^{0}$

What are angles in a circle ?

A circle has a total of 360 degrees all the way around the center, so if that central angle determining a sector has an angle measure of 60 degrees, then the sector takes up 60/360 or 1/6, of the degrees all the way around.

a. m < A = $\frac{1}{2}$ mBD

= $\frac{1}{2}$ × $78^{0}$

= $39^{0}$

b. mCE = 2m < CBE

= 2 × $20^{0}$

= $40^{0}$

c. m < c = m < A = $39^{0}$

d. m < D = $\frac{1}{2}$ mAC

= $\frac{1}{2}$ × $74^{0}$

= $37^{0}$

e. m < ABE = $\frac{1}{2}$ mAC + $\frac{1}{2}$ mCE

= $37^{0}$ + $20^{0}$

= $57^{0}$

Hence, m < A =  $39^{0}$ , mCE =  $40^{0}$, m < c = $39^{0}$, m < D = $37^{0}$ m < ABE = $57^{0}$

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