Answer:
∠DAC = ∠ACE ( By Alt.Int.∠s )
x = ∠ACE
Then, ∠ECF = 3x
∠ADB + ∠ADC = 180°
( ∠s on Straight line have sum 180°)
72° + ∠ACD = 180°
∠ADC = 180° - 72°
∠ADC = 108°
∠ACD + ∠ACF + ∠ECF = 180°
108° + x + 3x = 180°
4x = 180° - 108°
x = 72° / 4
Hope it helpful to you
Step-by-step explanation:
AE || EC
On Line BDC
On Line DCF
x = 18°
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Answer:
AE || EC
∠DAC = ∠ACE ( By Alt.Int.∠s )
x = ∠ACE
Then, ∠ECF = 3x
On Line BDC
∠ADB + ∠ADC = 180°
( ∠s on Straight line have sum 180°)
72° + ∠ACD = 180°
∠ADC = 180° - 72°
∠ADC = 108°
On Line DCF
∠ACD + ∠ACF + ∠ECF = 180°
( ∠s on Straight line have sum 180°)
108° + x + 3x = 180°
4x = 180° - 108°
x = 72° / 4
x = 18°
Hope it helpful to you
Verified answer
Answer:
Step-by-step explanation:
AE || EC
∠DAC = ∠ACE ( By Alt.Int.∠s )
x = ∠ACE
Then, ∠ECF = 3x
On Line BDC
∠ADB + ∠ADC = 180°
( ∠s on Straight line have sum 180°)
72° + ∠ACD = 180°
∠ADC = 180° - 72°
∠ADC = 108°
On Line DCF
∠ACD + ∠ACF + ∠ECF = 180°
( ∠s on Straight line have sum 180°)
108° + x + 3x = 180°
4x = 180° - 108°
x = 72° / 4
x = 18°