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in Triangles by (38.0k points)
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In Fig., AB || DE. Find ∠ACD.

1 Answer

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Since,

AB ‖ DE

∠ABC = ∠CDE (Alternate angles)

∠ABC = 40°

In ΔABC

∠A + ∠B + ∠ACB = 180°

30° + 40° + ∠ACB = 180°

∠ACB = 180° – 70°

= 110° (i)

Now,

∠ACD + ∠ACB = 180°(Linear pair)

∠ACD + 110° = 180°[From (i)]

∠ACD = 180° – 110°

= 70°

Hence, ∠ACD = 70°.

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