Two radio antennas are 120 m apart on a north-south line, and they radiate in phase at a frequency of 3.4 MHz. All radio measurements are made far...

Let's start by finding the wavelength of the radio waves radiating from the antennas. It will be

`lambda = c/f = (3*10^8)/(3.4*10^6) m = 88.2 m`

The constructive interference will occur when the path difference is an integer multiple of the wavelength. The path difference is the sum of the lengths of the segments `A_1P` and `A_2P` (please see the attached image.) Note also that `A_1P = A_2P` is the point P is on the East-West line passing midway between the two antennas, and that `A_1M = A_2M = 120/2 = 60 m` .

If the path difference is one wavelength:

`A_1P + 2A_2P = 2A_1P = lambda` , then

`A_1P = lambda/2 = 44.1` meters. This is shorter than distance `A_1M` , and thus impossible (hypotenuse in the right triangle has to be longer than an other side.) The next possible path difference resulting in constructive difference is twice the wavelength:

`2A_1P = 2lambda` . Then,

`A_1P = lambda = 88.2 m` .

The angle measured north of east is the angle labeled `alpha ` on the attached image. It can be found from

`sin(alpha) = (A_1M)/(A_1P) = 60/88.2 = 42.8` , or about 43 degrees.

This would correspond to choice B. However, this angle would get smaller as one goes further away from the antennas, so it is the largest possible angle at which the constructive interference occurs (not the smallest possible angle.) The angle `MA_1P` is the one that would get larger. This angle is equal to the one measured East of North, and it equals 90 - 43 = 47 degrees, which would then be choice C. So it seems that there is a discrepancy in which angle the problem is asking for.