42°
Since DE || AC, we have ∠DBC = ∠CBA = 84° (corresponding angles).
Therefore, in triangle EDB, ∠EDB + ∠DBC + ∠BED = 180°54° + 84° + ∠BED = 180°∠BED = 42°
In triangle BCA, ∠BCA + ∠CAB + ∠ABC = 180°∠BCA + 84° + 54° = 180°∠BCA = 180° - 84° - 54°∠BCA = 42°
Thus, ∠BCA = 42°.
42°
Since DE || AC, we have ∠DBC = ∠CBA = 84° (corresponding angles).
Therefore, in triangle EDB, ∠EDB + ∠DBC + ∠BED = 180°
54° + 84° + ∠BED = 180°
∠BED = 42°
In triangle BCA, ∠BCA + ∠CAB + ∠ABC = 180°
∠BCA + 84° + 54° = 180°
∠BCA = 180° - 84° - 54°
∠BCA = 42°
Thus, ∠BCA = 42°.