Answer
$R = 50.9~\Omega$
Work Step by Step
In part (a), we found that $~~Z = 64.0~\Omega$
It is given that the current leads the emf by $0.650~rad$
Therefore, $\phi = -0.650~rad$
We can find an expression for $X_L - X_C$:
$tan~\phi = \frac{X_L-X_C}{R}$
$X_L-X_C = R~tan~\phi$
We can find the resistance $R$:
$Z = \sqrt{R^2+(X_L-X_C)^2} = 64.0~\Omega$
$\sqrt{R^2+(R~tan~\phi)^2} = 64.0~\Omega$
$\sqrt{R^2~(1+~tan^2~\phi)} = 64.0~\Omega$
$R^2~(1+~tan^2~\phi) = (64.0~\Omega)^2$
$R^2 = \frac{(64.0~\Omega)^2}{1+~tan^2~\phi}$
$R = \frac{64.0~\Omega}{\sqrt{1+~tan^2~\phi}}$
$R = \frac{64.0~\Omega}{\sqrt{1+~tan^2~(-0.650~rad)}}$
$R = 50.9~\Omega$
