Answer:
[tex]28^\text{o}[/tex]
Step-by-step Explanation:
[tex]\text{1. }\angle \text{ADC}=\angle\text{ABC}=62^\text{o}\ \ \ [\text{Inscribed angles on same arc are equal.}]\\\\\text{2. }\angle\text{BCA}=90^\text{o}\ \ \ [\text{Inscribed angle on a semicircle is }90^\text{o}.]\\[/tex]
[tex]\text{3. }\angle \text{BCA}+\angle\text{BAC}+\angle\text{ABC}=180^\text{o}\ \ \ [\text{Sum of angles of triangle is }180^\text{o}.]\\\text{or, }90^\text{o}+\angle \text{BAC}+62^\text{o}=180^\text{o}\\\text{or, }\angle \text{BAC}=28^\text{o}[/tex]
