急~F.5 MATHS

=.=.=.=.=.=.=.=.=要咁樣先提交到=.=?

Let <EBD=x , <BED=y
<EDN=70° (< in alt.segment)
x+y+70°=180°
x = 110°-y ~1

<EDC=x (< in alt.segment)
<CED=180-y (adj. < on st.line)
50°+x+(180-y) =180°
x-y= -50 ~2

Put 1 into 2,we have
(110-y) -y =-50
y=80°

Put y =80° into 1,we have
x= 110-80
x=30°

SO <EDC =30°

∠BDE = ∠ABE = 70 deg (∠ in alt. segment)
Let ∠EDC be x.
∠DBE = ∠EDC = x (∠ in alt. segment)
In ΔBDE,
∠DBE + ∠BDE + ∠DEB = 180 deg (∠ sum of Δ)
x + 70 deg + ∠DEB = 180 deg
∠DEB = 110 deg - x
In ΔCDE,
∠EDC + ∠DCE = ∠DEB (ext. ∠ of Δ)
x + 50 deg = 110 deg - x
2x = 60 deg
x = 30 deg
Hence, ∠EDC = 30 deg

(angle)BDE=(angle)ABE=70 (angle in alt. segment)
Let (angle)CDE=x
(angle)EBD=(angle)WDC (angle in alt. segment)
=x
(angle)CDB+(angle)BCD+(angle)CBD=180 (angle sum of triangle)
(angle)EDC+(angle)EDB+(angle)BCD+(angle)CBD=180
x+70+50+x=180
2x=60
x=30
(angle)EDC=30



(sorry but the symbol of degree and angle can't be typed)