Here's my approach to this problem......first note....the minute hand must be ahead of the hour hand, initially......if it wasn't, the clock would have to run backwards in order for the hour hand to switch positions with the minute hand !!!!
Let the initial angle between the hands = x......and note that in two hours the hour hand has closed this gap by pi/3 rads......so....... we can let the remaining angle [in rads] that the hour hand has to close to reach the minute hand's original position just be θ.......and it closes this distance in some number of minutes after two hours
Meanwhile.......the minute hand returns to its original position after two hours. And to reach the original position of the hour hand it must travel 2pi - ( pi/3 + θ ) = [ 5pi/3 - θ] rads....and it travels this distance in the same time that the hour hand takes to travel θ rads
So......we need to find out how far each hand travels [ in rads] in one minute.......for the minute hand, this is just [ 2pi] / 60 = pi/30 rads.....and for the hour hand.....this distance is just [ pi/6 ] / 60 = pi/360 rads
So......Distance traveled/ rate = time and we can equate the times thusly
Distance the hour hand travels in rads in some minutes after two hours/ the number of rads it moves in one minute =
Distance that the minute hand travels is the same number of minutes after two hours/ the rads it moves in one minute
And we have :
θ/ [pi/360] = [ 5pi/3 - θ] / [pi/30] simplify
360θ = 30 [5pi/3 - θ ]
360θ = 50pi - 30θ
390θ = 50 pi
39θ = 5 pi
θ = 5pi/39 and this is the number of rads that the hour hand travels in the minutes after two hours
So....to find the number of minutes, we have
[5pi/39] / [ pi/360] =
[360 * 5] / 39 ≈ 46.15 minutes ≈ 46 minutes and 9 seconds
So....the man is gone for 2 hrs, 46 minutes and 9 seconds [ approximately]
