Question

HELP PLZ
21. Find the perimeter of the triangle:
2
A
4715

Answer 1

Answer:

16$\sqrt{5}$+4$\sqrt{15}$ 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$\sqrt{15}$/hypotenuse

hypotenuse=$\frac{4\sqrt{15} }{\frac{\sqrt{3} }{2} }$

hypotenuse=8($\frac{\sqrt{15} }{\sqrt{3} }$)

hypotenuse=8$\sqrt{5}$

So the left and right hand sides of the larger triangle are both 8$\sqrt{5}$ respectively.

To find the base, once again consider the smaller triangle, and this time, use pythagorus:

a²=b²+c²

(8$\sqrt{5}$)²=(4$\sqrt{15}$)²+c²

320=240+c²

80=c²

c=4$\sqrt{15}$

So the perimeter is 8$\sqrt{5}$+8$\sqrt{5}$+4$\sqrt{15}$=16$\sqrt{5}$+4$\sqrt{15}$=51.26902