Answer
$84.26^{\circ}$
Work Step by Step
Let, the phase constant of the $7.0\;mm$ wave is $\phi$.
The phase constant of the $5.0\;mm$ wave is $0$.
Therefore, the amplitude of the resultant wave is given by
$A=\sqrt {(5+7\cos\phi)^2+(7\sin\phi)^2}\;mm$
According to the given condition
$A=9\;mm$
or, $\sqrt {(5+7\cos\phi)^2+(7\sin\phi)^2}=9$
or, $(5+7\cos\phi)^2+(7\sin\phi)^2=81$
or, $49(\cos^2\phi+\sin^2\phi)+70\cos\phi+25=81$
or, $70\cos\phi=7$
or, $\phi=\cos^{-1}(0.1)$
or, $\phi=84.26^{\circ}$
$\therefore$ The phase constant of the $7.0\;mm$ wave is $84.26^{\circ}$.
