I have a question regarding "Professional No Limit Hold'Em: Vol 1" specifically Hand No. 2 and 3 in the chapter called "Fundamentals in Practice" on page 87.

Hand No. 2. You have $100 on the button in a $1-$2 game. An
aggressive player who has $800 opens for $6, another player calls,
and you call with 66. The blinds fold, and the flop comes
754, giving you a pair and an open-ended straight draw. The
pot is $21.

Your aggressive opponent bets $20, and the next player folds.
You have $94 remaining. This flop is excellent for your hand,
giving you a nearly 1 in 3 chance of making a straight by the river.
(By the Four Times Rule it's 32 percent.) Plus, you could draw
out on a bigger pair by catching a set, and you may even have the
hest hand.

Its the quote above... I don't understand how the authors calculate the pot to be 61$,
when the pot on the flop was 21$ + the 20$ bet from our aggro opponent?

Any answer will be greatly appreciated

typically, when calculating a raise size, you first include how much you'd have to call first and then the raise on top of that, so it is 21 + 20 + your 20 and then the raise is 74. This is convenient to be able to easily see the pot odds you'd be giving your opponent, in this case he would need to call 74 into a 61 pot, so he needs around 35% equity to continue

as a side note, 66 should just fold preflop here if the blinds are even remotely competent

Thanks for the reply I just find it easier to say: 21 + 20 + 94 = 135. And he has to call 74. 135:74=1.8:1=35%