In ABC, A=50^ , B=70^ and bisector of C meets AB in D. Find the angles of the triangles ADC and BDC.

We know that the sum of all three angles of a triangle is equal to 180^ Therefore, for the given ABC, we can say that: A+ B+ C=180^ (Sum of angles of ABC) 50^ +70^ + C=180^ C=180^ -120^ C=60^ ACD= BCD= C 2 (CD bisects C and meets AB is D.) ACD= BCD=60^ 2 =30^ Using the same logic for the given ACD, we can sat that: DAC+ ACD+ ADC=180^ 50^ +30^ + ADC=180^ ADC=180^ -80^ ADC=100^ If we use the same logic for the given BCD, we can say that: DBC+ BCD+ BDC=180^ 70^ +30^ + BDC=180^ BDC=180^ -100^ BDC=80^ Thus, For ADC: A=50^ , D=100^ , C=30^ For BDC: B=70^ , D=80^ , C=30^

In ABC, A=50^ , B=70^ and bisector of C meets AB in D. Find the a…

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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 $BDC.$

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Answer

We know that the sum of all three angles of a triangle is equal to $180^\circ $
Therefore, 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 $\triangle \text{BDC}:\angle \text{B}=70^\circ, \ \angle \text{D}=80^\circ, \ \angle \text{C}=30^\circ$