In the given figure, ABC is a triangle in which ∠A : ∠B : ∠C = 3 : 2 : 1
Question:

In the given figure, ABC is a triangle in which ∠A : ∠B : ∠C = 3 : 2 : 1 and AC ⊥ CD. Find the measure of ∠ECD.

 

Solution:

Let $\angle A=(3 x)^{\circ}, \angle B=(2 x)^{\circ}$ and $\angle C=x^{\circ}$

From $\triangle A B C$, we have:

$\angle A+\angle B+\angle C=180^{\circ} \quad$ [Sum of the angles of a triangle]

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

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

$\Rightarrow x=30^{\circ}$

$\begin{aligned} \therefore & \angle A=3(30)^{\circ}=60^{\circ} \\ & \angle B=2(30)^{\circ}=60^{\circ} \end{aligned}$

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

Side BC of triangle ABC is produced to E.

$\therefore \angle A C E=\angle A+\angle B$

$\Rightarrow \angle A C D+\angle E C D=(90+60)^{\circ}$

$\Rightarrow 90^{\circ}+\angle E C D=150^{\circ}$

 

$\Rightarrow \angle E C D=60^{\circ}$

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