a seesaw consists of a plank 4.5m long which is supported by a pivot at its center and moves in a vertical plane above the pivot. if the height of the pivot pillar above the ground is 1m, through what maximum angle can the seesaw beam move?

Answer 1

Did you make a sketch ?
I see an isosceles triangle on its side with equal sides 4.5 and a vertical line of 1
label the top of the plank P, the fulcrum F
draw a horizontal to hit the height at Q
PFQ is a right-angled triangle with hypotenuse 4.5, and a vertical leg of .5
angle QFP is half of the angle we want.
tan QFP = .5/4.5 = 5/45 = 1/9
angle QFP = appr 6.34°
So the beam moves through an angle of 12.68°

Answer 2

we could have just used the cosine law

1^2 = 4.5^2 + 4.5^2 - 2(4.5)(4.5)cosØ
1 = 20.25 + 20.25 - 40.5cosØ
40.5 cosØ = 39.5
cosØ = 39.5/40.5 = .9753
Ø = 12.759°

And I just realized that in my first solution I should have used the sine instead of tangent

sin QFP = .5/4.5 = 5/45 = 1/9
angle QFP = appr 6.3794°
So the beam moves through an angle of 12.759°
just like above