A neutral pi meson, rest mass 135 MeV, decays symmetrically into two photons while moving at high speed. the energy of each photon in the lab system is 100 MeV. find the angle 'a' in the lab system between the momentum of each photon and the initial line of motion. the answer in the book is 42 degrees, but i found that it's 39.88 or something like that, here's my solution the initial four momentum is: (E/c,P)=(2E_ph,P1+P2) now by using lorentz invariant here: M_pi_rest^2*c^2=(E/c)^2-P^2=(4E_ph^2/c^2-(P1+P2)^2) and we know that E_ph/c=P1=P2 so we get (2E_ph^2/c^2)-(E_ph/c)^2*cos(a)=M_pi^2c^2 from there i calclulated the angle, but didn't get 42 degrees, perhaps the book is outdated and this accounts for the rounding up. any advice as to wether i got it confused or the book?