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