someone solve the b of 3 plz plz plz..urgent
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∠ABD is an exterior angle of ∆DBC
∠ABD = ∠BCD + ∠BDC ..( remote interior angle theorem)
= 89° + 27°
= 116°
∠ABD = ∠ABE ......(B-E-D)
In ∆ ABE
∠ABE + ∠BAE + ∠AEB = 180° ( sum of measures of angles of the triangle is 180°)
22°+116°+∠AEB = 180°
138° + ∠AEB = 180°
∠AEB = 42°
∠AEB + ∠AED = 180 (angles in linear pair)
42° + c =180°
c = 180°-42°
c = 138°
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