An aerialist on a high platform holds onto a trapeze attached to a support by an 8.0-m cord. (See the drawing.) Just before he jumps off the platform, the cord makes an angle of 41° with the vertical. He jumps, swings down, then back up, releasing the trapeze at the instant it is 0.75 m below its initial height. Calculate the angle θ that the trapeze cord makes with the vertical at this instant.

Answer 1

Imagine two triangles:
1) ΔABC : point A (where the cord is attached ), point B (where the aerialist holds the trapeze at initial position), point C (the intersection of the vertical line from A and horizontal line from B),
2) ΔADM: point M (where the aerialist holds the trapeze at its left position), point D (the intersection of the vertical line from A and horizontal line from M)
AC = AB•cosθₒ =L•cosθₒ,
AD =AM•cosθ = L•cosθ,
CD = d.
AD =AC +CD,
L•cosθ = L•cosθₒ +d.
cosθ =( L•cosθₒ +d)/L =
=(8•cos41º+0.75)/8 = 0.848.
arcos 0.848 = 32º.

Answer 2

arccos 0.848 = 32º.