Two radio antennas separated by d 297 m as shown in the fig

Two radio antennas separated by d = 297 m, as shown in the figure below, simultaneously transmit identical signals of the same wavelength. A radio in a car traveling due north receives the signals.

(a) If the car is at the position of the second maximum, what is the wavelength of the signals?
  
m

(b) How much farther must the car travel to encounter the next minimum in reception? (Hint: Determine the path difference between the two signals at the two locations of the car.)
  m

Solution

let the left above point is A , below point is B

and car is at C and later at D

a.) Path difference = BC - AC = ( 550² + 1000² )^1/2   - ( 250² + 1000²  )^1/2  = 1141.271221 - 1030.776406

= 110.4948146 m

This should be equal to 2 time the wavelength to be hearing the 2nd maximum after the zeroth, i.e central maximum.

So, 2 = 110.4948146

= 55.247 m

b.) For the next minimum, the path difference should be 2.5

so, 2.5 = BD - AD

2.5 x 55.247 = ( (y +150)² + 1000² )^1/2 - ( (y-150)² + 1000²   )^1/2

138.117 = ( y² + 300y + 1022500)^1/2 - ( y² - 300y + 1022500 )^1/2  

138.117 + ( y² - 300y + 1022500 )^1/2 =  ( y² + 300y + 1022500)^1/2

Solving this equation gives, y = 480.221 m

So, the car has to further move a distance of 480.221 - 400 = 80. 221 m