Mathematics

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

$64^o$

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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Single Correct Medium Published on 09th 09, 2020
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