1.

In the given fig. DE || BC, ∠ADE = 70° and ∠BAC = 50°, then angle ∠BCA = ___

Answer»

By converse of Thale’s theorem DE II BC 

∠ADE = ∠ABC = 70° 

Given ∠BAC = 50° 

∠ABC + ∠BAC + ∠BCA = 180° (Angle sum prop of triangles) 

70° + 50° + ∠BCA = 180° 

∠BCA = 180° - 120° = 60°



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