The lecturer with a monotone voice of 440 Hz and you are both 2m away from the long, smooth and tall wall. You will hear sound that comes directly to you from the lecturer along with sound that comes back from the wall on it’s way to reaching you. You need the path here needs to have the same angle onto the wall as it has after reflection. How far from the lecturer will you pick your seat so you can’t hear the voice at all and find your well-deserved rest?

the difference in the path length between the direct and reflected sound should be an odd half wavelength to facilitate cancellation

... 1/2, 3/2, 5/2...

the speed of sound is 340 m/s

so the wavelength is 340/440 m

the diagram is an isosceles triangle with an altitude of 2 m
... the base of the triangle is the direct path