Question

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

$\angle ECD.$

Answer 1

In $\Delta ABC,\angle A:\angle B:\angle C$ =3:2:1

BC is produced to D and CE $\perp$ AC

$\therefore \angle A+\angle B+\angle C={180}^{\circ }$ (Sum of angles of a triangles)

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

$\therefore 3x+2x+x={180}^{\circ }\Rightarrow 6x={180}^{\circ }$

$\Rightarrow x=\frac{{180}^{\circ }}{6}={30}^{\circ }$

$\therefore \angle A=3x=3\times {30}^{\circ }={90}^{\circ }$

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

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

In $\Delta ABC$,

Ext. $\angle ACD=\angle A+\angle B$

$\Rightarrow {90}^{\circ }+\angle ECD={90}^{\circ }+{60}^{\circ }={150}^{\circ }$

$\therefore \angle ECD={150}^{\circ }-{90}^{\circ }={60}^{\circ }$