Mathematics

In a $$\triangle ABC$$, it is given that $$\angle A : \angle B : \angle C = 3 : 2 : 1$$ and $$CD \perp AC$$. Find $$\angle ECD$$.


SOLUTION
In a $$\triangle ABC$$, it is given that
$$\angle A : \angle B : \angle C = 3 : 2 : 1$$

It can also be written as

$$\angle A = 3x, \angle B = 2x$$ and $$\angle C = x$$

We know that the sum of all the angles in triangle $$ABC$$ is $$180^{\circ}$$.

$$\angle A + \angle B + \angle C = 180^{\circ}$$

By substituting the values we get

$$3x + 2x + x = 180^{\circ}$$

By addition

$$6x = 180^{\circ}$$

By division

$$x = 180/6x = 30^{\circ}$$

Now by substituting the value of $$x$$ we get

$$\angle A = 3x = 3 (30^{\circ}) = 90^{\circ}$$

$$\angle B = 2x = 2 (30^{\circ}) = 60^{\circ}$$

$$\angle C = x = 30^{\circ}$$

We know that in the triangle $$ABC$$ exterior angle is equal to the sum of two opposite interior angles

So we can write it as

$$\angle ACE = \angle A + \angle B$$

By substituting the values we get

$$\angle ACE = 90^{\circ} + 60^{\circ}$$

By addition

$$\angle ACE = 150^{\circ}$$

We know that $$\angle ACE$$ can be written as $$\angle ACD + \angle ECD$$

So we can write it as

$$\angle ACE = \angle ACD + \angle ECD$$

By substituting the values we get

$$150^{\circ} = 90^{\circ} + \angle ECD$$

It is given that $$CD \perp AC$$ so $$\angle ACD = 90^{\circ}$$

On further calculation

$$\angle ECD = 150^{\circ} - 90^{\circ}$$

By subtraction

$$\angle ECD = 60^{\circ}$$

Therefore, $$\angle ECD = 60^{\circ}$$.
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