Answer
The air temperature is $-20.7132$ $^{\circ}$F.
Work Step by Step
1. Write the wind chill temperature (WCT) formula from exercise $37$ $WCT=\begin{cases}T, 0\leq v \leq 3\\35.74 + 0.6215T − 35.75v^{0.16} + 0.4275T v^{0.16}, 3 < v\end{cases}$
2. Wind speed $v=48$mi/h.
since, $v>3$
$WCT=35.74 + 0.6215T − 35.75v^{0.16} + 0.4275T v^{0.16}$
3. Now substitute values of wind speed and WCT in the formula.
$-60=35.74 + 0.6215T − 35.75\times 48^{0.16} + 0.4275T\times 48^{0.16}$
4. Isolate the terms with $T$ and solve for $T$
$0.6215T + 0.4275T\times 48^{0.16}=-60 + 35.75\times 48^{0.16} - 35.74$
$\implies T (0.6215 + 0.4275\times 48^{0.16})=-29.3237$
$\implies T(1.4157)=-29.3237$
$\implies T=-20.7132$
Hence, the air temperature is $-20.7132$ $^{\circ}$F.
