Question

If angle < akl and < bkl are supplementry and m < akl= 3z+20 what is < akl and < bkl.

Answer 1

Answer:

given that<akl=3z+20

let the<bkl=z

supplementryangle=180

so' 3z+20+ z=180

4z+20=180

4z=180-20

4z =160

z=160/4

z=40

so' <akl=3z+20

40×3+20

120+20

140

Answer 2

Answer:

badalalala

Step-by-step explanation:badalala