Question
In $\triangle \text{ABC}, \angle \text{A}=50^\circ, \angle \text{B}=70^\circ$ and bisector of $\angle \text{C}$ meets AB in D. Find the angles of the triangles ADC and BDC.
✓
Answer
We know that the sum of all three angles of a triangle is equal to 180ºTherefore, for the given $\triangle \text{ABC},$ we can say that:
$\angle \text{A}+\angle \text{B}+\angle \text{C}=180^\circ$ $($Sum of angles of $\triangle \text{ABC})$
$\Rightarrow 50^\circ+70^\circ+\angle \text{C}=180^\circ$
$\angle \text{C}=180^\circ-120^\circ$
$\angle \text{C}=60^\circ$
$\angle \text{ACD}=\angle \text{BCD}=\frac{\angle \text{C}}{2}$ (CD bisects $\angle \text{C}$ and meets AB is D.)
$\Rightarrow \angle \text{ACD}=\angle \text{BCD}=\frac{60^\circ}{2}=30^\circ$
Using the same logic for the given $\triangle \text{ACD},$ we can sat that:
$\angle \text{DAC}+\angle \text{ACD}+\angle \text{ADC}=180^\circ$
$\Rightarrow 50^\circ+30^\circ+\angle \text{ADC}=180^\circ$
$\angle \text{ADC}=180^\circ-80^\circ$
$\angle \text{ADC}=100^\circ$
If we use the same logic for the given $\triangle \text{BCD},$ we can say that:
$\angle \text{DBC}+\angle \text{BCD}+\angle \text{BDC}=180^\circ$
$\Rightarrow 70^\circ+30^\circ+\angle \text{BDC}=180^\circ$
$\angle \text{BDC}=180^\circ-100^\circ$
$\angle \text{BDC}=80^\circ$
Thus,
For $\triangle \text{ADC}:\angle \text{A}=50^\circ, \ \angle \text{D}=100^\circ, \ \angle \text{C}=30^\circ$
For $\triangle \text{BDC}:\angle \text{B}=70^\circ, \ \angle \text{D}=80^\circ, \ \angle \text{C}=30^\circ$
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