Mathematics

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

$60^o$

##### 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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Single Correct Medium Published on 09th 09, 2020
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