MCQ
In $\triangle\text{ABC},$ $\angle\text{A}=50^{\circ},$$\angle\text{B}=70^{\circ}$ and bisector of $\angle\text{C}$ Meets $AB$ in $D$ figure. Measure of $\angle\text{ADC}$
is.
is.
- A$50^\circ $
- ✓$100^\circ $
- C$30^\circ $
- D$70^\circ $
✓
Answer
Correct option: B.
$100^\circ $
In $\triangle\text{ADC},$
$\angle\text{ADC}+\angle\text{DAC}+\angle\text{ACD}=180^{\circ}$ [angle sum property of a triangle]
$\Rightarrow \ \angle\text{ADC}+50^{\circ}+\angle\text{ACD}=180^{\circ}$ $[\because\angle\text{DAC}=50^{0}]$
$\Rightarrow \ \angle\text{ACD}=130^{\circ}-\angle\text{ACD}......(\text{i})$
In $\triangle\text{ DBC},$ $\angle\text{ADC}=\angle\text{DBC}+\angle\text{BCD}$
$[\because$ exterior angle is equal to sum of opposite interior angles$]$
$\Rightarrow \ \angle\text{ADC}=70^{\circ}+\angle\text{ACD}$ $[\because\angle\text{ACD}=\angle\text{BCD}]$
$\Rightarrow \ \angle\text{ADC}=70^{\circ}+130^{\circ}-\angle\text{ADC}$ [from equation $(i)$]
$\Rightarrow \ \angle\text{ADC}=200^{\circ}-\angle\text{ADC}$
$\Rightarrow 2\ \angle\text{ADC}=200^{\circ}$
$\Rightarrow \ \angle\text{ADC}=\frac{200^{\circ}}{2}$
$\Rightarrow \ \angle\text{ADC}=100^{\circ}$
$\angle\text{ADC}+\angle\text{DAC}+\angle\text{ACD}=180^{\circ}$ [angle sum property of a triangle]
$\Rightarrow \ \angle\text{ADC}+50^{\circ}+\angle\text{ACD}=180^{\circ}$ $[\because\angle\text{DAC}=50^{0}]$
$\Rightarrow \ \angle\text{ACD}=130^{\circ}-\angle\text{ACD}......(\text{i})$
In $\triangle\text{ DBC},$ $\angle\text{ADC}=\angle\text{DBC}+\angle\text{BCD}$
$[\because$ exterior angle is equal to sum of opposite interior angles$]$
$\Rightarrow \ \angle\text{ADC}=70^{\circ}+\angle\text{ACD}$ $[\because\angle\text{ACD}=\angle\text{BCD}]$
$\Rightarrow \ \angle\text{ADC}=70^{\circ}+130^{\circ}-\angle\text{ADC}$ [from equation $(i)$]
$\Rightarrow \ \angle\text{ADC}=200^{\circ}-\angle\text{ADC}$
$\Rightarrow 2\ \angle\text{ADC}=200^{\circ}$
$\Rightarrow \ \angle\text{ADC}=\frac{200^{\circ}}{2}$
$\Rightarrow \ \angle\text{ADC}=100^{\circ}$
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