我想問條數學題

AB = AC, therefore, triangle ABC is an isos. triangle, so angle ACB also 30 degree.
Therefore, angle CAB = 180 - 30 - 30 = 120.
Therefore, area of triangle = (1/2)(2)(2)sin120. = 2sin120 = sqrt3........(1)
By cosine rule,
BC^2 = 2^2 + 2^2 - 2(2)(2)cos120 = 4 + 4 + 4 = 12. BC = sqrt12 = 2sqrt3.
Let radius of the incribed circle = r.
Therefore, area of triangle = (2r)/2 + (2r)/2 + (2rsqrt3)/2 = r + r + rsqrt3
= r(2 + sqrt3).......(2)
From (1) and (2), r(2 + sqrt3) = sqrt3
r = sqrt3/(2 + sqrt3).

2008-08-20 11:18:40 補充：
sqrt3/(2 + sqrt3) can be simplified to 2sqrt3 - 3.