I need help in this applied maths question

a particle of mass 10kg is pulled over horizintal ground.the pulling force has magnitude 100N and makes an angle A with the horizontal.where tan A=3/4. find the normal reaction and the acceleration

A is arctan(3/4)

The horizontal component of the pulling force is 100*cos(arctan(3/4)) = 100*(4/5) = 80
F = ma
80 = 10a
a = 80/10
a = 8 m/(s^2)

The vertical component of the pulling force is 100*sin(arctan(3/4)) = 100*(3/5) = 60
Assuming the particle is near the Earth's surface, the gravitational field strength is 9.8, so the weight of the particle is 10 * 9.8 = 98N
So the normal force applied by the ground onto the object is 98 - 60 = 38N

tan A = Perpendicular/Base
Complete a right triangle to get sin A = Perpendicular / Hypotenuse = 4/5

Balance forces in vertical direction
N = mg - FsinA
N = 10 x 10 - 100 x 3/5 = 100 - 60 = 40N

Balance forces in horizontal direction

FcosA = ma
Complete a right triangle to get cos A = Base / Hypotenuse = 4/5
a = 100 x 4/5 / 10 = 8 m/s^2

Hope it helped :)