a) tanb = 2√3 = cotc
    cotb = 1/2√3 = tanc
cotb = 1/2√3 = sinb/cosb
=> sinb = 1 = cosc
     cosb = 2√3 = sinc
b) tanb = 2√3 = ah/bh = 2√3/bh 
=> bh = 1
=>ab = √(1^2+(2√3)^2) = √13
hd = ah^2/bh = 12
bc = 12 + 1 = 13
ac = √13^2 - 13) = 2√39