Mathematics

In the given figure; $$ AE\parallel BD $$,  $$ AC\parallel ED $$ and $$ AB = AC $$. Find $$ \angle a  $$,


ANSWER

$$64^o$$


SOLUTION
$$\angle EDC = 58^{\circ}$$ (Vertically opposite angles)

Given $$AE \parallel CD$$
$$c = \angle EDC = 58^{\circ}$$ (Corresponding angles)

Also, $$AC \parallel ED$$
$$\angle ACB = 58^{\circ}$$ (Corresponding angles)

Since, $$AB = AC$$
$$\angle ABC = \angle ACB$$ (Isosceles triangle property)
$$b = \angle ABC = 58^{\circ}$$ (Corresponding angles)

Now, In $$\triangle ABC$$,
$$\angle ABC + \angle ACB + \angle BAC = 180 $$
$$58 + 58 + \angle BAC = 180$$
$$\angle BAC = 180 - 116$$
$$\angle BAC = 64^{\circ}$$
$$a = 64^{\circ}$$
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