Question

plz answer this 2...... u will get 98 points and will be marked as the braniest............
plz plz begging you guys................

Answer 1

the answer for first question is c) 20 degree ;
from fig.
6y+2y+y=180
9y=180
y=20
6y=20×6=120
2y=20×2=40

the answer for second question is beam 50 degree 130 degree ;
from fig.
PAB=ABC-vertically opp. angle=50
ABC+BCD=180-alternate angles
50+X=180
x=180-50
x=130
BCD=CDT-alternate angles
Hope this helps
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Answer 2

1)………………………………………
Given, 3 angles: 6y,y, and 2y

6y+y+2y=180° (angles on a straight line)
9y=180°
y=180/9°
y=20°

So, Answer : (c) 20°


2)………………………………………
Given,
\_PAB = 50°
To find,
\_a and \_b
\_ABC=\_PAB( alternate angles)
\_ABC=50°
a=50°
Now,
\_ABC+\_BCD=180°( Co-interior angles)
50°+b=180°
b=180°-50°
b=130°

So, answer : (b) 50°,130°