Answer:
16[tex]\sqrt{5}[/tex]+4[tex]\sqrt{15}[/tex] or 51.26902
Step-by-step explanation:
Consider one of the smaller, right-angled triangles within the main triangle, and use sine to find the hypotenuse of the small triangles:
sinA=opposite/hypotenuse
sin60=4[tex]\sqrt{15}[/tex]/hypotenuse
hypotenuse=[tex]\frac{4\sqrt{15} }{\frac{\sqrt{3} }{2} }[/tex]
hypotenuse=8([tex]\frac{\sqrt{15} }{\sqrt{3} }[/tex])
hypotenuse=8[tex]\sqrt{5}[/tex]
So the left and right hand sides of the larger triangle are both 8[tex]\sqrt{5}[/tex] respectively.
To find the base, once again consider the smaller triangle, and this time, use pythagorus:
a²=b²+c²
(8[tex]\sqrt{5}[/tex])²=(4[tex]\sqrt{15}[/tex])²+c²
320=240+c²
80=c²
c=4[tex]\sqrt{15}[/tex]
So the perimeter is 8[tex]\sqrt{5}[/tex]+8[tex]\sqrt{5}[/tex]+4[tex]\sqrt{15}[/tex]=16[tex]\sqrt{5}[/tex]+4[tex]\sqrt{15}[/tex]=51.26902
