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Square

In a pentagon...

Question

In a pentagon $A B C D E, A B$ is parallel to $D C$ and $∠ A : ∠ E : ∠ D = 3 : 4 : 5$ . Find $∠ E$ in degrees.

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120

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170

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180

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Solution

The correct option is D 120
Given, $A B ∥ C D$
and BC is intersecting them. So, sum of their interior angles are supplementary.
$∴ ∠ B + ∠ C = 180$
As we Know, sum of angles of a Pentagon is 540,
$∴ ∠ A + ∠ B + ∠ C + ∠ D + ∠ E = 540$
$∠ A + ∠ D + ∠ E + 180 = 540 (∵ ∠ B + ∠ C = 180) o r, ∠ A + ∠ D + ∠ E = 540 - 180 o r, ∠ A + ∠ D + ∠ E = 360$
Given , $∠ A : ∠ E : ∠ D = 3 : 4 : 5$
Let the angle be $x$ ,
Therefore, $∠ A = 3 x, ∠ E = 4 x, ∠ D = 5 x$
$∠ A + ∠ D + ∠ E = 360$ ,
$o r, 3 x + 4 x + 5 x = 360 o r, 12 x = 360 o r, x = \frac{360}{12} = 30 N o w, ∠ E = 4 x = 4 \times 30 = 120$

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