Question

The side BC of $\Delta ABC$ is produced to a point D. The bisectors of $\angle A$ meet side BC in L. If $ABC={30}^{\circ }$ and $\angle ACD={115}^{\circ }$,

then $\angle ALC$=

Answer 1

In $\Delta ABC,BC$ is produced to D $\angle B={30}^{\circ },\angle ACD={115}^{\circ }$

AL is bisector of $\angle A$

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

$\Rightarrow {115}^{\circ }+\angle A={30}^{\circ }$

$\therefore \angle A+{115}^{\circ }-{30}^{\circ }={85}^{\circ }$

and $\angle C={180}^{\circ }-{115}^{\circ }={65}^{\circ }$

and $\angle CAL=\frac{1}{2}\angle A$ $(\because$ AL is bisectors)

=$\frac{1}{2}\times {85}^{\circ }=42\frac{1_{\circ }}{2}$

In $\Delta ALC,$

Ext. $\angle ACD=\angle ALC+\angle CAL$

${115}^{\circ }=\angle ALC+42\frac{1_{\circ }}{2}$

$\therefore \angle ALC={115}^{\circ }-42\frac{1_{\circ }}{2}={72.5}^{\circ }$