2.2cos(X)/2.2sin(x)+9.81=0.16029

Question

An open rectangular tank contains water up to about half of its depth. This tank accelerates at a=2.20 m/s² up a slope of α (alpha), which causes the free water surface to form an angle θ (theta) with the original horizontal plane (Figure 5). What is the α angle of the slope?

 

Note - I have work out by trail and error that the angle is approx 36 degress

Answer 1

me assume the formula yu giv shood be

{[2.2*kosine(x)] / [2.2*sine(x)]} +9.81=0.16029

or {...}=0.16029-9.81=-9.64971

{...} become 1/tan(x) or kotan(x) or rerite it tu tan(x)=1/(-9.64971)

tan(x)=-0.103630057

invers tan giv x=-5.91644595 deg