Question

please answer fastly​

Answer 1

Answer:

(i) ACB = 120 °

(ii) ACD = 60 °

(iii) AEB = 150 °

Step-by-step explanation:

(i) In ∆ CAD

angle CAD + angle ADC = angle ACB (exterior angle property)

So, 65° + 55° = angle ACB

angle ACB = 120°

(ii) In ∆ CAD

angle CAD + angle ADC + angle ACD = 180° ( angle sum property of a triangle)

65° + 55° + ACD = 180°

ACD = 180 - 120

angle ACD = 60 °

(iii) In ∆ BCE

angle CBE + angle BCE = angle AEB (exterior angle property)

So, 30° + 120° = angle AEB

angle AEB = 150°

Answer 2

Answer:

there is 180 degree in a triangle .

So, Angle A =65°

Angle B=55°

Sum of Angle A and B =(65+55)°= 120°

Angle ACD = 180°-120°=60°

Angle ACD + Angle ACB= 180 °

Angle ACB = 180° - 60° = 120°

Angle BEC+ Angle B + Angle ACB = 180°

Angle BEC + 30° +120° = 180°

Angle BEC+ 150° = 180°

Angle BEC = 180°-150°= 30°

Angle BEC+ Angle AEB = 180°

30° + Angle AEB= 180°

Angle AEB = 180° - 30° =150°

Hence , (I) Angle ACB = 120°

Angle(ii) ACD= 60°

ACD= 60°Angle (iii)AEB= 150°

Hope it helps