Answer
$Φ$ $\approx$ $63 °$
Work Step by Step
$m$ = $\frac{Δy}{Δx}$ = $tanΦ$
$0$ $\lt$ $Φ$ $\lt$ $\frac{\pi}{2}$
$f(x)$ = $x^{2}$
$f'(x)$ = $2x$
So the slop of the tangent to the curve at the point (1,1) is 2
$m$ = $tanΦ$
$2$ = $tanΦ$
$Φ$ = $tan^{-1}2$
$Φ$ $\approx$ $63 °$