CTET 2 Math and Science 07 Jul 2024 Paper

In a triangle, the angle bisector of an angle divides the opposite side into segments that are proportional to the adjacent sides.
Calculation:
Triangle Sum Property: The sum of the interior angles of a triangle is always ${180}^{\circ }$. For triangle $△\mathit{A}\mathit{B}\mathit{C}$, we have:
$\Rightarrow \angle \mathit{A}+\angle \mathit{B}+\angle \mathit{C}={180}^{\circ }$
$\Rightarrow \angle \mathit{C}={180}^{\circ }-{50}^{\circ }-{70}^{\circ }$
$\Rightarrow \angle \mathit{C}={60}^{\circ }$
Angle Bisector Property:
Since CD is the bisector of $\angle \mathit{A}\mathit{C}\mathit{B}$, it divides $\angle \mathit{C}$ into two equal parts.
Thus, $\angle \mathit{A}\mathit{C}\mathit{D}=\angle \mathit{B}\mathit{C}\mathit{D}=\angle \mathit{C}∕2={60}^{\circ }∕2={30}^{\circ }$
Angle Calculation at Point D:
In $△\mathit{A}\mathit{D}\mathit{C}$, we need to find $\angle \mathit{A}\mathit{D}\mathit{C}$.
The sum of angles in $△\mathit{A}\mathit{D}\mathit{C}$ is also ${180}^{\circ }$.
$\Rightarrow \angle \mathit{A}+\angle \mathit{A}\mathit{C}\mathit{D}+\angle \mathit{A}\mathit{D}\mathit{C}={180}^{\circ }$
Substitute the known values: ${50}^{\circ }+{30}^{\circ }+\angle \mathit{A}\mathit{D}\mathit{C}={180}^{\circ }$
$\Rightarrow \angle \mathit{A}\mathit{D}\mathit{C}={180}^{\circ }-{80}^{\circ }={100}^{\circ }$