plzz help me........
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Since AB||DC
Therefore, x=∠BDC(alt.angles)
>x=35°
In ΔABD,
x+y+∠DAB=180(angle sum property of a triangle)
>35°+80°+y=180°
>115°+y=180°
>y=180°-115°
>y=65°
In ΔDCB,
y-30°=65°-30°=35°
Since, ∠BDC+∠DCB+∠CBD=180°
>35°+z+35°=180°
>z+70°=180°
>z=180°-70°
>z=110°
Hence,
x=35°
y=65°
z=110°
Hope This Helps!
Therefore, x=∠BDC(alt.angles)
>x=35°
In ΔABD,
x+y+∠DAB=180(angle sum property of a triangle)
>35°+80°+y=180°
>115°+y=180°
>y=180°-115°
>y=65°
In ΔDCB,
y-30°=65°-30°=35°
Since, ∠BDC+∠DCB+∠CBD=180°
>35°+z+35°=180°
>z+70°=180°
>z=180°-70°
>z=110°
Hence,
x=35°
y=65°
z=110°
Hope This Helps!
SANAM01:
x= 35 how do you get it
DC and BA are parallel and BD is the transversal, therefore the alternate angles CDB and DBA are equal,hence x is 35°
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