At a playground, a child slides down a slide that makes a 42° angle with the horizontal direction. the coefficient of kinetic friction for the child sliding on the slide is 0.20. what is the magnitude of her acceleration during her sliding?

Assuming metric units, metre, kilogram and seconds

Best approach: draw a free body diagram and identify forces acting on the child, which are:
gravity, which can be decomposed into normal and parallel (to slide) components 
N=mg(cos(theta)) [pressing on slide surface]
F=mg(sin(theta))  [pushing child downwards, also cause for acceleration]
   m=mass of child (in kg)
   g=acceleration due to gravity = 9.81 m/s^2
   theta=angle with horizontal = 42 degrees

Similarly, kinetic friction is slowing down the child, pushing against F, and equal to
Fr=mu*N=mu*mg(cos(theta))
  mu=coefficient of kinetic friction = 0.2

The net force pushing child downwards along slide is therefore
Fnet=F-Fr
=mg(sin(theta))-mu*mg(cos(theta))
=mg(sin(theta)-mu*cos(theta))   [ assuming sin(theta)> mu*cos(theta) ]

From Newton's second law,
F=ma, or 
a=F/m
=mg(sin(theta)-mu*cos(theta))  /  m
= g(sin(theta)-mu*cos(theta))   [ m/s^2]

In case imperial units are used, g is approximately 32.2 feet/s^2.
and the answer will be in the same units [ft/s^2] since sin, cos and mu are pure numbers.

40 & 50 in the right corner are the complementary angles as when placed in a sum (added together) they make 90°