The drawing shows a loudspeaker A and point C, where a listeer is positioned. A second loudspeaker B is located somewhere to the right of A. Both speakers vibrate in phase and are playing a 68.6-Hz tone. The speed of sound is 343 m/s. What is the closest to speaker A that speaker B can be located, so that the listener hears no sound? Picture: First, I connected C and B to make a triangle. I then bisect the triangle with a line straight down from C to make 2 triangles (with the extra point being D). I call AB line = x and CD line = y. Next, I set up the destructive interference where the point in which line AB has to be 1 wavelength and line CB is half wavelength. So, the difference between AC and CB has to be CB - CA = (n*wavelength)/2 for smallest distance, n would have to be 1. with this, I list all the known and unknowns which are AD = AC*cos(60) DB = AB-AD -> x-cos(60) CD = AC*sin(60) CB = sqrt(sin^2(60)+ (x-cos^2(60))) So then: wavelength/2 = sqrt(sin^2(60)+cos^2(60) + x^2 - 2*cos(60x)) I solve to get: x^2 - (2*cos(60)x) + (1 + wavelength^2/4) use quadratic equation to get x = 2.845 and x = -1.845 The correct answer is 3.89 m. So, I don't know if I'm missing something here. I hope you guys can help. Thanks.