Answer:
The perimeter of ∆ABC is [tex]P=14.35\ units[/tex]
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
When a circle has an inscribed angle that "cuts out" a semi-circle, like the one in the attached figure, then the inscribed angle is a right angle. so
m∠C=90°
The perimeter of triangle ABC is equal to
[tex]P=AB+BC+AC[/tex]
[tex]AB=2r=2(3)=6\ units[/tex] ----> is equal to the diameter of the circle
Find the length side AC
[tex]cos(35\°)=\frac{AC}{AB}[/tex]
[tex]AC=cos(35\°)(AB)[/tex]
substitute
[tex]AC=cos(35\°)(6)=4.91\ units[/tex]
Find the length side BC
[tex]sin(35\°)=\frac{BC}{AB}[/tex]
[tex]BC=sin(35\°)(AB)[/tex]
substitute
[tex]BC=sin(35\°)(6)=3.44\ units[/tex]
Find out the perimeter
[tex]P=AB+BC+AC[/tex]
substitute
[tex]P=6+3.44+4.91=14.35\ units[/tex]
