Given that HC is the diameter of the circle, then:
![\hat{HC}=180\degree](https://img.qammunity.org/2023/formulas/mathematics/college/tvjbec1v46769y4ef2uufmdy0mymdmcrwk.png)
From the diagram, HC can be expressed as follows:
![\hat{HC}=\hat{HA}+\hat{AB}+\hat{BC}](https://img.qammunity.org/2023/formulas/mathematics/college/q5ggbinn9tkdvf2n4ovbrusvkownvfkesm.png)
Substituting with HC = 180°, HA = 83°, and BC = 50°, and solving for AB:
![\begin{gathered} 180\degree=83\degree+\hat{AB}+50\degree \\ 180\degree-83\degree-50\degree=\hat{AB} \\ 47\degree=\hat{AB} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/22wqkmasbeuqxh2p82edwdk6chx0yel2ww.png)
The relation between the angle outside the circle, ∠1, and the intersected arcs AB and HD are:
![m\angle1=(1)/(2)(\hat{HD}-\hat{AB})](https://img.qammunity.org/2023/formulas/mathematics/college/lbfbqbs31ni1v9tuqc9lc8e8vc9y49gz2i.png)
Substituting with HD = 135°, and AB = 47°, we get:
![\begin{gathered} m\angle1=(1)/(2)(135\degree-47\degree) \\ m\angle1=44\degree \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/fxf1n3ih94oau1zzzxp08k8tqv7x5eh4ak.png)