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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