In the figure, $AC \perp CE$ and $\angle A: \angle B : \angle C = 3:2:1$, find the value of $\angle ECD$."

Given:

In the given figure, $AC \perp CE$ and $\angle A: \angle B : \angle C = 3:2:1$.

To do:

We have to find the value of $\angle ECD$.

Solution:

We know that,

Sum of the angles in a triangle is $180^o$.

$\angle A + \angle B + \angle C = 180^o$

Let $\angle A = 3x$

This implies,

$\angle B = 2x$ and $\angle C = x$

Therefore,

$3x + 2x + x = 180^o$

$6x = 180^o$

$x = \frac{180^o}{6}$

$x = 30^o$

This implies,

$\angle A = 3x = 3(30^o) = 90^o$

$\angle B = 2x = 2(30^o) = 60^o$

$\angle C = x = 30^o$

In $\triangle ABC$,

External $\angle ACD = \angle A + \angle B$

$90^o + \angle ECD = 90^o + 60^o = 150^o$

$\angle ECD = 150^o-90^o = 60^o$

The value of $\angle ECD$ is $60^o$.

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Updated on: 10-Oct-2022

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