Find the measure of all the angles and all the arcs in the figure below.Please

Answer:
• mAD=60°
,• mBF=60°
,• mCF=60°
,• m∠COD=120°
,• m∠B=90°
,• m∠C=30°
,• m∠D=60°
,• m∠DAB=120°
Explanation:
The line AC is the diameter of the circle.
[tex]\begin{gathered} m\widehat{AC}=180\degree \\ m\widehat{AC}=m\widehat{AD}+m\widehat{DC} \\ 180\degree=m\widehat{AD}+120\degree \\ m\widehat{AD}=180\degree-120\degree \\ m\widehat{AD}=60\degree \end{gathered}[/tex]Similarly, line DF is a diameter, thus:
[tex]\begin{gathered} m\widehat{DF}=180\degree \\ m\widehat{DF}=m\widehat{AD}+m\widehat{AB}+m\widehat{BF} \\ 180\degree=60\degree+60\degree+m\widehat{BF} \\ m\widehat{BF}=180\degree-120\degree \\ m\widehat{BF}=60\degree \end{gathered}[/tex]In like manner, using line DF:
[tex]\begin{gathered} m\widehat{DF}=180\degree \\ m\widehat{DF}=m\widehat{DC}+m\widehat{CF} \\ 180\degree=120\degree+m\widehat{CF} \\ m\widehat{CF}=180\degree-120\degree \\ m\widehat{CF}=60\degree \end{gathered}[/tex]Angle COD is the central angle subtended by arc CD at the centre.
[tex]\begin{gathered} m\angle\text{COD}=m\widehat{CD} \\ m\angle\text{COD}=120\degree \end{gathered}[/tex]Angle B is the angle subtended by the diameter AC at the circumference of the circle. The angle in a semicircle is 90 degrees, therefore:
[tex]m\angle B=90\degree[/tex]Angle C is the angle subtended by arc AB at the circumference.
[tex]\begin{gathered} m\widehat{AB}=2\times m\angle C \\ 60\degree=2\times m\angle C \\ m\angle C=\frac{60\degree}{2} \\ m\angle C=30\degree \end{gathered}[/tex]Next, we find the measure of angle D.
In Triangle AOD,
[tex]\begin{gathered} m\angle O=\text{mAD}=60\degree \\ OD=OA(\text{radi}i) \\ \triangle\text{AOD is Isosceles} \\ \implies\angle O+2\angle D=180\degree \\ 60\degree+2m\angle D=180\degree \\ 2m\angle D=120\degree \\ m\angle D=60\degree \end{gathered}[/tex]Finally, we find the measure of angle DAB.
[tex]\begin{gathered} m\angle\text{DAB}=m\angle\text{DAO}+m\angle\text{CAB} \\ =60\degree+60\degree \\ =120\degree \end{gathered}[/tex]