Joel selected an acute angle $x$ (strictly between 0 and 90 degrees) and wrote the values of $\sin x$, $\cos x$, and $\tan x$ on three different cards.  Then he gave those cards to three students, Malvina, Paulina, and Georgina, one card to each, and asked them to figure out which trigonometric function (sin, cos, or tan) produced their cards.  Even after sharing the values on their cards with each other, only Malvina was able to surely identify which function produced the value on her card.  Compute the sum of all possible values that Joel wrote on Malvina's card.
The functions $\sin x,$ $\cos x,$ $\tan x$ are one-to-one on the interval $(0^\circ,90^\circ).$  Since Malvina could deduce her function, the value of $x$ can also be deduced.  In particular, $\sin x,$ $\cos x,$ and $\tan x$ are all known.  Since they cannot deduce Paulina's function and Georgina's function, their values must be equal.

If $\sin x = \cos x,$ then $\tan x = 1,$ so $x = 45^\circ.$  Then Malvina's value is 1.

If $\sin x = \tan x = \frac{\sin x}{\cos x},$ then $\cos x = 1.$  But $\cos x$ cannot achieve 1 on the interval $(0^\circ,90^\circ).$

If $\cos x = \tan x = \frac{\sin x}{\cos x},$ then $\sin x = \cos^2 x = 1 - \sin^2 x.$  Then
\[\sin^2 x + \sin x - 1 = 0.\]By the quadratic formula,
\[\sin x = \frac{-1 \pm \sqrt{5}}{2}.\]Since $-1 \le \sin x \le 1,$
\[\sin x = \frac{-1 + \sqrt{5}}{2}.\]This is the case where $\cos x = \tan x,$ so Malvina's value is $\sin x = \frac{-1 + \sqrt{5}}{2}.$

Therefore, the sum of the possible numbers on Malvina's card is
\[1 + \frac{-1 + \sqrt{5}}{2} = \boxed{\frac{1 + \sqrt{5}}{2}}.\]