Perimeter of Triangle ABC 
I'm stuck in question iii. Can somebody give me some hints? How could ABC be triangle?
 A: As for how $ABC$ can be a triangle,

A: Notice, $\triangle ABC$ is a right triangle & $\theta=\pi/3=60^\circ$, $AC=2r=16\ cm$ $$AB=2\times 8\sin30^\circ=8\ cm$$
hence, in right $\triangle ABC$,  $$BC=\sqrt{AC^2-AB^2}=\sqrt{16^2-8^2}=8\sqrt 3$$
hence the perimeter of right triangle $ABC$ $$=AB+BC+AC=8+8\sqrt 3+16$$ $$=\color{red}{24+8\sqrt 3}$$
A: Connect B to A and B to C with straight lines. 
In this case, it may help to drop a perpendicular down from B and note that because you know the angle, you could compute the height of B and then work with a base and height to compute the area of a triangle formed by connecting points A,B and C together.
A: I've found the answer by using Law of Cosines.
$BC^2=8^2+8^2-2(8)(8)\cos 120=192$
$BC=8\sqrt{3}$
$AB+BC+AC=24+8\sqrt{3}$
A: Use the cosine theorem in triangle $AOB$ to calculate the length of $AB$:
$AB^2=AO^2+OB^2-2AO\cdot OB\cdot\cos\theta$
Since $AO$ and $OB$ are radius they both equal 8, so the calculation is straightforward.
Repeat the same procedure with the triangle $BOC$ to find the length of $BC$ (notice that the angle $\widehat{BOC}=\pi-\theta$).
