Answer:
[tex]P=(7\sqrt{15}+7\sqrt{5})\ units[/tex]
Step-by-step explanation:
see the attached figure to better understand the problem
step 1
Find the length side AB (hypotenuse)
we know that
[tex]cos(30^o)=\frac{AC}{AB}[/tex]
substitute
[tex]cos(30^o)=\frac{7\sqrt{5}}{AB}[/tex]
Remember that
[tex]cos(30^o)=\frac{\sqrt{3}}{2}[/tex]
so
[tex]\frac{\sqrt{3}}{2}=\frac{7\sqrt{5}}{AB}[/tex]
[tex]AB=\frac{14\sqrt{5}}{\sqrt{3}}[/tex]
[tex]AB=\frac{14\sqrt{15}}{3}\ units[/tex]
step 2
Find the length side BC
[tex]sin(30^o)=\frac{BC}{AB}[/tex]
[tex]sin(30^o)=\frac{1}{2}[/tex]
[tex]\frac{1}{2}=\frac{BC}{AB}[/tex]
[tex]BC=\frac{AB}{2}[/tex]
substitute the value of AB
[tex]BC=\frac{\frac{14\sqrt{15}}{3}}{2}[/tex]
[tex]BC=\frac{7\sqrt{15}}{3}\ units[/tex]
step 3
Find the perimeter
[tex]P=AB+BC+AC[/tex]
substitute
[tex]P=\frac{14\sqrt{15}}{3}+\frac{7\sqrt{15}}{3}+7\sqrt{5}[/tex]
[tex]P=\frac{21\sqrt{15}}{3}+7\sqrt{5}[/tex]
[tex]P=(7\sqrt{15}+7\sqrt{5})\ units[/tex]
