Ninth Grade Physics Question (Best Answer Will Be Chosen)

A skier of mass 82.4 kg comes down a slope
of constant angle 40 with the horizontal.
What is the force on the skier parallel to the
slope? The acceleration of gravity is 9.8m/s2 .
Answer in units of N

What force normal to the slope is exerted by
the skis?
Answer in units of N

Since there is no mention of friction, I assume it to be zero:

Fp = m*g*sinΘ = 519.1 N

Fn = m*g*cosΘ = 618.6 N

parallel: a = g*cos(theta)
a = 9.8 * cos(40)
a = 7.5072355425659847449834479754431
F = ma
F = 82.4 * 7.5072355425659847449834479754431
F = 618.59620870743714298663611317651
F = 618.6N

Normal: a = g*sin(theta)
a = 9.8*sin(40)
a = 6.2993185749280853979619054170912
F = 519.06385057407423679206100636831
F = 519.1N

1) formula is f=mgsin(teta) 2) formula is f=mgcos(teta) same values as before
where g=9.8m/s2 f= 82.5*9.8*.76=614.46N
teta=40 degrees
m= mass of skier
hence f=82.4*9.8*sine(40)
f=82.4*9.8 *0.64=516.8128N