You can also do it using trig.
sin(\angle EDC) = \frac{8}{10}
\angle EDC = sin^{-1}\frac{4}{5}
\angle ADE = 90 - \angle EDC
\angle ADE = 90 - sin^{-1}\frac{4}{5}
Cos(90 - sin^{-1}\frac{4}{5}) = \frac{AD}{6}
AD = 6Cos(90 - sin^{-1}\frac{4}{5})
AD = 4.8
You now have 2...
1.C
2.B
3.B
4.C
5.B
6.C
7.C
8.A
9.B
10.B (wasn't too sure about this, I couldn't remember what the independent variable was..)
11.D
12.D
13.A
14.C
15.C
16.D (Both B and D seemed to makes sense for this question so I'm not sure..)
17.E
18.B
19.D
20.D
Re: Wow, fuck that
I found question 5 to be a lot harder then any other question. All that stuff for part a and it was only 1 mark each... And I couldn't make any sense out of part b ii.