Trong tam giác vuông \(ABC\), ta có:
\(\eqalign{
& AB = BC.\sin C = BC.\sin {30^0} = 4.{1 \over 2} = 2\left( {dm} \right) \cr
& AC = BC.\cos C = BC.\cos {30^0} = 4.{{\sqrt 3 } \over 2} \cr&= 2\sqrt 3 \left( {dm} \right) \cr} \)
Ta có: \(S_{xq}= πRl = π. 2. 4 = 8 π \) \((dm^2).\)
\(\displaystyle V = {1 \over 3}\pi {R^2}h = {1 \over 3}\pi {.2^2}.2\sqrt 3 = {{8\sqrt 3 .\pi } \over 3}(d{m^3}).\)
