Toboggan down hill, FIND ANGLE?

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Homework Statement



A girl starting from rest, toboggans down a hill of height 15m. The mass of the girl and toboggan is 42kg. The speed of the girl and toboggan at bottom of hill is 2.9m/s. If force of friction is 112 N, what is the slope of the hill? Ignore air resistance.

Homework Equations



KE=PE, 15sin(theta), F=mg

The Attempt at a Solution



I started by equating initial and final energies by saying KE final is equal to PE initial plus the force of friction. (.5mv)squared = mgh - (mu)Fn
I ended up with 0.1429=sin(theta) which equals 8.2 degrees. The answer was 16 degrees, where did I go wrong?
 
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What is 15sin([tex]\vartheta[/tex]) meant to be?
 
I am trying to figure out the distance down the slope that she travels so I can use that along with the height (15m) to find the angle.
 
15sin(theta) will not give you the distance traveled down the slope.
 
Well I suppose I am stuck then.
 
You know what to do, you just have your trig relations mixed up a bit. If sintheta = height/hypotenuse, then 15sintheta = height x height/hypotenuse...
 
I should know this, but that still leaves me with unknowns for theta and hypotenuse.
I am drawing a blank on how to find the hypotenuse with what is given.
Thanks for the replies by the way.
 
Sintheta = height/hypotenuse where hypotenuse = the distance traveled down the slope. Since we know the height, we can state the hypotenuse in terms of the height and theta, that is hypotenuse = height/sintheta. This way you can use 15m/sintheta in the expression for the work done by friction, and then solve for theta, which was your idea to begin with. You just misused the sin ratio!
 
W=Ff*d


5997.4/112N=d

d=53.6m
 
So how did you determine W was 5997.4?
 
I have to disagree with Squeezebox, the work done is not as quoted. To calculate the work done, you can apply conservation of energy.