Given that HC is the diameter of the circle, then:
[tex]\hat{HC}=180\degree[/tex]
From the diagram, HC can be expressed as follows:
[tex]\hat{HC}=\hat{HA}+\hat{AB}+\hat{BC}[/tex]
Substituting with HC = 180°, HA = 83°, and BC = 50°, and solving for AB:
[tex]\begin{gathered} 180\degree=83\degree+\hat{AB}+50\degree \\ 180\degree-83\degree-50\degree=\hat{AB} \\ 47\degree=\hat{AB} \end{gathered}[/tex]
The relation between the angle outside the circle, ∠1, and the intersected arcs AB and HD are:
[tex]m\angle1=\frac{1}{2}(\hat{HD}-\hat{AB})[/tex]
Substituting with HD = 135°, and AB = 47°, we get:
[tex]\begin{gathered} m\angle1=\frac{1}{2}(135\degree-47\degree) \\ m\angle1=44\degree \end{gathered}[/tex]
