Find the value of x

<AEC =X°

<EAB = 116°

the by (consecutive)

180 = <EAB + <1

<1 = (180 - 116 )

<1=. 64°

<ECD = 124°

then ,180 = <ECD + <2

<2= (180 - 124)

<2= 56°

= Now add <1 + <2

= (64+ 56)°

➡x° = 120°

Solution given:

Construction:

EF parallel to AB and CD is drawn.

let suppose <AEF=a

and

<CEF=b

now

<EAB+<EAB=180°[being co interior angle]

116°+a=180°

a=180°-116°=64°

again

<DCE +<CEF=[being co interior angle]

124°+b=180°

b=180°-124°

b=56°

Now

we know

x=a+b=64°+56°=120°

Value of x is 120°

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