Iam confused on number 5 and 6! ? it's do really soon, so as soon as possible! ?

5)

First we need to find the height which will be the length of the two sides of the triangle on the right...

Using the Law of Sines...(sina/A=sinb/B=sinc/C for any triangle)

h/sin60=8√6/sin90   (and since sin90=0)

h=8√6(sin60)

h=(8√6√(3/4)  (sin60=√(3/4))

h=√((64*6*3)/4)

h=√288

...

Since x is the hypotenuse of the right isosceles triangle...

x^2=2h^2, and using h found above we have:

x^2=2*288

x^2=576

x=24 units...

6)

The sides of the right isosceles triangle on the right are:

5^2=2s^2

2s^2=25

s^2=25/2

s=√(25/2)

And this side length is opposite the 60° angle of the triangle on the left.  Again using the law of sines we can find x.

x/sin90=√(25/2)/sin60   again sin90=1 and sin60=√(3/4) so

x=√(25/2)/√(3/4)  which is equal to:

x=√(25/2)√(4/3)

x=√(50/3)

x=5√(2/3)...rationalizing the denominator by multiply by √(3/3) gives you:

x=5√2√3/3

x=(5√6)/3

Teri ma ki chhut‌‌‌‌‌‌

Um what? rephrase that ?