Answer:
The part A can be calculate using the secant theorem below
To figure out the value of triangle ADB, we will use the formula below
![\begin{gathered} \angle ADB=(1)/(2)(ARC\text{ AB-ARC AC}) \\ ARC\text{ AB=146}^0 \\ \text{ARC AC=50}^0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/w2nto1rbqg1j91l60cffpjul40vidf8paz.png)
By substituing the values, we will have
![\begin{gathered} \begin{equation*} \angle ADB=(1)/(2)(ARC\text{ AB-ARC AC}) \end{equation*} \\ \angle ADB=(1)/(2)(146^0-50^0) \\ \angle ADB=(1)/(2)(96^0) \\ \angle ADB=48^0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hpvowxfczpp69b98zya2bxk6ivqdrgc1io.png)
Hence,
The value of angle ADB is
![\Rightarrow\angle ADB=48^0](https://img.qammunity.org/2023/formulas/mathematics/college/zzjfj5h9ootkak3cgwya1xtavaexk0yvxb.png)
Part B:
To figure out the value of arc angle CD, we will use the formula below
Hence,
The formula will be
![\begin{gathered} \angle AEB=(1)/(2)(ARC\text{ AB+ARC CD}) \\ ARC\text{ AB=41}^0 \\ \angle AEB=42^0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vvni7f67fayyoi17kxqdznd0mh252gzyz5.png)
By substituting the values, we will have
![\begin{gathered} \begin{equation*} \angle AEB=(1)/(2)(ARC\text{ AB+ARC CD}) \end{equation*} \\ 42^0=(1)/(2)(41^0+arc\text{ CD}) \\ cross\text{ multiply, we will have} \\ 84=41^0+arc\text{ CD} \\ arc\text{ CD=84}^0-41^0 \\ arc\text{ CD=43}^0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/fta8vnh0g4vg9xvubzrcsq8si74ot9lur5.png)
Hence,
The value of arc CD is
![\Rightarrow arc\text{ CD=43}^0](https://img.qammunity.org/2023/formulas/mathematics/college/jzmjmkjwn9q77wdvbdjn6vd8dwhd8kk96t.png)