1. The easiest way to do this is simply the cosine rule:
a^2 = b^2 + c^2 - 2bccosA
= 2^2 + 1^2 - 2.2.1cos60
= 3
a = sqrt(3)
Another is to realise triangle ADE must be right-angled (according to my intuition, anyway, though I'd have to construct a proof to be sure - although the fact that Pythag works does prove it's right-angled anyway) and Pythagoras' Theorem can therefore be used: b^2 = c^2 - a^2 = 2^2 - 1^2 = 3 => b = sqrt(3).
EDIT: Presumably the "type" of triangle is scalene? It's neither equilateral or isosceles, so...
2. Drawing a line CO gives a triangle COB. CO splits the 60 degree angle BCA in half, making angle BCO 30 degrees, meaning angle CBO 30 degrees too, and angle COB 120 degrees.
Using the sine rule: 2/sin120 = BO/sin30 => BO = 2/sqrt(3) => BD = 4/sqrt(3) as BO is the radius and BD is the diameter.
3. Length DF is 1, which can be figured out either by realising that triangles AED and DFC are similar, meaning their proportions are the same, or by realising that EF acts as a diagonal for parallelogram BEDF and so BE = DF. Through similar logic it can be shown that angle EDF is 60 degrees.
Area of a triangle = 1/2.a.b.sinC = 1/2.2.1.sin60 = sqrt(3)/2 cm^2
I fucking hate geometry questions.
