Answer:
a
 [tex]\Phi  =4.524 \  rad[/tex]
b
  [tex]I  =  0.40637 I_o [/tex]
Explanation:
From the question we are told that
 The distance of separation is  [tex]d =  9.0 \  m[/tex]
 The frequency is [tex]f = 120 \  MHz  =  120  *10^{6} \  Hz[/tex]
  The distance of the receiver from the antennas is [tex]D =  150 \  m[/tex]
  The intensity measured is  [tex]I =  I_o[/tex]
  The change in position of the receiver is by  [tex]\Delta D =  1.8 \  m[/tex]
Gnerally the phase difference is mathematically represented as
    [tex]\Phi  = \frac{ 2 \pi  *  \Delta D}{\lambda}[/tex]
Here [tex]\lambda[/tex] Â is the wavelength which is mathematically represented as
    [tex]\lambda  =  \frac{c}{f}[/tex]
Here c is the speed of light with value  [tex]c = 3.0 *10^{8} \  m/s[/tex]
=>   [tex]\lambda  =  \frac{3.0 *10^{8}}{ 120  *10^{6} }[/tex]
=>   [tex]\lambda  =  \frac{3.0 *10^{8}}{ 120  *10^{6} }[/tex]
=>   [tex]\lambda  = 2.5 \  m[/tex]
So
 [tex]\Phi  = \frac{ 2* 3.142  * 1.8 }{2.5}[/tex]
 [tex]\Phi  = \frac{ 2* 3.142  * 1.8 }{2.5}[/tex]
 [tex]\Phi  =4.524 \  rad[/tex]
Generally the intensity measured by the receiver is Â
   [tex]I  = I_o  cos^2 [\frac{\Phi}{2} ][/tex]
=>  [tex]I  = I_o  [cos [\frac{4.524}{2} ]]^2[/tex]
=>  [tex]I  = I_o  [cos [ 2.262]]^2[/tex]
=> Â [tex]I Â = Â 0.40637 I_o [/tex]
