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The first question is an application of the law of sines to find a side of the right-hand triangle. You must then recognize that since two of the angles of the triangle on the right are equal, their opposite sides must be equal.

The second question was really similar in the way they wanted you to go about solving it.
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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

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