Question

In $△ABC,\angle A={50}^o,\angle B={70}^o$ and bisector $\angle C$ meets $AB$ in $D$. Find the angles of the triangles $ADC$ and $BDC$.

Answer 1

In $\begin{array}{c}ABC\\ \angle A+\angle B+\angle C=180\\ {50}^{\circ }+70+\angle C=180\\ {120}^{\circ }+\angle C=180\\ \angle C={180}^{\circ }-{120}^{\circ }={60}^{\circ }\end{array}$

Since CD is the bisector $\angle C$

So, $\angle ACD=\angle BCD=\frac{\angle C}{2}=\frac{{60}^{\circ }}{2}={30}^{\circ }$

In triangle $ADC$

$\begin{array}{c}\angle ACD+\angle DAC+\angle ADC={180}^{\circ }\\ {30}^{\circ }+{50}^{\circ }+\angle ADC={180}^{\circ }\\ \angle ADC={180}^{\circ }-{80}^{\circ }\\ \angle ADC={100}^{\circ }\\ \angle ADC+\angle BDC={180}^{\circ }\\ {100}^{\circ }+\angle BDC={180}^{\circ }\\ \angle BDC={180}^{\circ }-{100}^{\circ }\\ \angle BDC={80}^{\circ }\end{array}$

So in triangle $ADC$

$\angle DAC={50}^{\circ },\angle ACD={30}^{\circ },\angle ADC={100}^{\circ }$

In triangle $BDC$

$\angle DBC={70}^{\circ },\angle BCD={30}^{\circ },\angle BDC={80}^{\circ }$

And $\angle BDC={80}^{\circ }$