Mathematics

In the given figure, AB // CD, EH // BC, $$\angle BAC = 60^{\circ} and\ \angle DGH = 40^{\circ}$$. Find the measures of  $$\angle ACD$$.


ANSWER

$$60^o$$


SOLUTION
Let, extended BC ends at point M.

Given, $$\angle DGH = 40^{\circ}$$

Thus, $$\angle DCM = \angle DGH = 40^{\circ}$$ (Corresponding angles of parallel lines EH and BM)

Also, $$\angle ABC = \angle DCM = 40^{\circ}$$ (Corresponding angle of parallel lines CD and AB)

Now, in $$\triangle ABC$$,

$$\angle ABC  + \angle ACB + \angle BAC = 180^{\circ} $$

$$40^{\circ} + 60^{\circ} + \angle ACB = 180^{\circ} $$

$$\angle ACB = 80^{\circ}$$

Now, $$\angle ACB + \angle ACD + \angle DCM = 180^{\circ} $$

$$80^{\circ} + \angle ACD + 40^{\circ} = 180^{\circ} $$

$$\angle ACD = 60^{\circ}$$
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