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In △ABC, ∠ABC = 100°, ∠BAC = 35° and BD ⊥ AC meets side AC in D. If BD = 2 cm, find ∠C, and length DC.

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We know that the sum of all angles of a triangle is 180°

Therefore, for the given △ABC, we can say that:

∠ABC + ∠BAC + ∠ACB = 180°

100° + 35° + ∠ACB = 180°

∠ACB = 180° –135°

∠ACB = 45°

∠C = 45°

On applying same steps for the △BCD, we get

∠BCD + ∠BDC + ∠CBD = 180°

45° + 90° + ∠CBD = 180° (∠ACB = ∠BCD and BD parallel to AC)

∠CBD = 180° – 135°

∠CBD = 45°

We know that the sides opposite to equal angles have equal length.

Thus, BD = DC

DC = 2 cm

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