Single photon dissociation of bromoform using synchrotron radiation has been investigated by Fourier transform visible fluorescence spectroscopy (FTVIS). The photodissociation of bromoform in the 12-18 eV energy range results in several products, among which are the CH(A(2)Delta) and CH(B-2 Sigma) radicals. Vibrational and rotational state distributions of the CH(A(2)Delta) are determined from their fluorescence spectra. From the threshold photon energy above which emission from the CH(A(2)Delta) radicals is observed, the most likely process leading to CH(A) formation is CHBr3 -> CH + 3Br rather than CHBr3 -> CH + Br + Br-2. The rotational Boltzmann temperatures in the CH(A -> X) emission spectra for v' = 0 and v' = I range between 1570 and 3650 K, depending on the excitation photon energy. From the high rotational excitation, the results suggest that the mechanism for the loss of three bromine atoms is most likely sequential. A small negative emission anisotropy of the CH(A) radicals [(I-par - I-per)/(I-par + 2I(per)) = -0.024 +/- 0.005] is constant across the action spectrum; a small net absorption dipole of CHBr3 in the vacuum ultraviolet is parallel to the 3-fold symmetry axis of the CHBr3 molecule. The state distributions of the CH(A 2 A) radicals from multiphoton dissociation of bromoform using the 266 nm output (three photons) of a femtosecond laser (Boltzmann temperatures: T-v'=0(rot) = 4250 +/- 300 K; T-v'=1(rot) = 3100 +/- 550 K) are compared to those from the single photon dissociation results (Boltzmann temperatures: T-v'=1(rot) = 3650 +/- 150 K; T-v'(rot) = 2400 +/- 200 K) at the same total excitation energy under collision free conditions. The analysis of the CH(A) rotational populations shows hotter rotational populations for the femtosecond experiment, also suggesting sequential dissociation of the bromoform in the femtosecond experiment. The duration of the femtosecond laser pulse is approximately 180 fs, setting a limit on the time scales for the multiple dissociations.