Answer:
• mAD=60°
,
• mBF=60°
,
• mCF=60°
,
• m∠COD=120°
,
• m∠B=90°
,
• m∠C=30°
,
• m∠D=60°
,
• m∠DAB=120°
Step-by-step explanation:
The line AC is the diameter of the circle.
![\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}](https://img.qammunity.org/2023/formulas/mathematics/college/ohvvgnbk5w5n0unt3ho8i6395jab3ca4n9.png)
Similarly, line DF is a diameter, thus:
![\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}](https://img.qammunity.org/2023/formulas/mathematics/college/42qhjh4qfb8d1hv9o0agddd8qla44ueiqe.png)
In like manner, using line DF:
![\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}](https://img.qammunity.org/2023/formulas/mathematics/college/urdhyxvl4k4228to2cjy9p0gbevu7ssu0t.png)
Angle COD is the central angle subtended by arc CD at the centre.
![\begin{gathered} m\angle\text{COD}=m\widehat{CD} \\ m\angle\text{COD}=120\degree \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/f9mg8r87xynrz3bza59q5kcphs0vd1675s.png)
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:
![m\angle B=90\degree](https://img.qammunity.org/2023/formulas/mathematics/college/l7ob7pqmppwohq3iev1tw21d21idkggtzu.png)
Angle C is the angle subtended by arc AB at the circumference.
![\begin{gathered} m\widehat{AB}=2* m\angle C \\ 60\degree=2* m\angle C \\ m\angle C=(60\degree)/(2) \\ m\angle C=30\degree \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/x5x0yjf8dls23abj6gu3p0wt54fg8if680.png)
Next, we find the measure of angle D.
In Triangle AOD,
![\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}](https://img.qammunity.org/2023/formulas/mathematics/college/rl671u1oag43854xc420ga67f2uh5l2204.png)
Finally, we find the measure of angle DAB.
![\begin{gathered} m\angle\text{DAB}=m\angle\text{DAO}+m\angle\text{CAB} \\ =60\degree+60\degree \\ =120\degree \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7zsjjhtrq3utxzz4p39tgc3bhyxmg4dt4h.png)