Question

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Answer 1

Answer:

∠CAB+∠DBA=180°

∠DBA=180° −∠CAB

=180° −130°

=50°

(AC∥BD, co-int. ∠s are supp.)

DBG being a straight line, ∠DBG=180°

⇒ ∠DBA+∠ABF+∠FBG=180°

⇒ 50+∠ABF+60°

=180°

⇒ ∠ABF=180°−110

⇒x°=70°

Now, AE∥BF ⇒ ∠EAB+∠ABF=180°(co-int ∠s)

⇒ x+70°

=180°

⇒x=110°

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