Verified answer

pulling force parallel to slope = 70 cos θ

pulling force normal to slope = 70 sin θ

weight is 20x9.8 = 196 N

force normal to slope due to weight = 196 cos 14

force parallel to slope due to weight = 196 sin 14

as there is no acceleration, force of friction + force parallel to slope due to weight = pulling force parallel to slope

10 + 196 sin 14 = 70 cos θ

70 cos θ = 57.42

cos θ = 0.8202

θ = 34.9º

I am a little dubious of my own solution, hope some other responder can confirm or else argue!

In a coordinate system where the sled's path is horizontal, the gravitational force is "mg" as usual, but its along-slope component is mg sin(14) while its component normal to the slope is mg cos(14).  The girl's applied force has an along-slope component of (70N)*cos(theta), and its component normal to the slope is (70N)*sin(theta).  The frictional force is then 

(mu)*[mg cos(14) - (70 N)*sin(theta)], which we are told is 10 N.

As both (mu) and theta are unknown, some further relationship is needed.

If the girl's effort is barely compensating for the frictional force and the along-slope component of gravity, then

(70 N)*cos(theta) = 10 N + (mg)*sin(14).

This implies cos(theta) = (10 N + 47.42 N)/(70 N)

which would imply theta = 34.88 degrees.

Then the normal force is

190.18 N - (70N)*sin(34.88) = 150.15 N,

which would imply (mu) = 0.0666 ... seems possible.