To solve this problem we need to apply doppler equation,
Our values are:
[tex]f_0 = 910Hz\\c = 75m\\v=344m/s\\\theta= 12.9\°\\[/tex]
Doppler equation is given by,
[tex]f= \frac{v}{v-c} f_0[/tex]
Substituting,
[tex]f= \frac{344}{344-75}*910[/tex]
[tex]f = 1163Hz[/tex]
[tex]\lambda = \frac{344}{1163}[/tex]
Wavelenght is equal to,
[tex]\lambda = \frac{c}{f}[/tex]
[tex]\lambda = 0.2956[/tex]
We can now find the distance by,
[tex]d= \frac{1}{2}\frac{\lambda}{sin\theta}[/tex]
[tex]d=\frac{1}{2}\frac{0.2956}{sin(12.9)}[/tex]
[tex]d= 0.66m[/tex]
b) We can find at what angles would sound first cancel through the same equation of wavelenght,
[tex]\lambda = \lambda{c}{f} = \frac{344}{1163}[/tex]
[tex]d* sin\theta = \frac{1}{2}*\lambda[/tex]
[tex]\theta = arcsin( 0.5*(344/1163)/0.66)=12.94\°[/tex]
