A aerialist on a high platform holds on to a trapeze attached to a sup

[cmkc109]

Homework Statement

A aerialist on a high platform holds on to a trapeze attached to a support by a 8.2-m cord. Just before he jumps off the platform, the cord makes an angle α of 39.8deg with the vertical. He jumps, swings down, then back up, releasing the trapeze at the instant it is 0.64 m below its initial height. Calculate the angle θ that the trapeze cord makes with the vertical at this instant.

Homework Equations

cos(theta) = A/H

The Attempt at a Solution

So I found the solution online.
8.2 cos 39.8 = 6.3
6.3 + 0.64 =6.94
cos ^-1 (6.94 / 8.2) = 32.2 deg

However I am not understanding it at all. I don't understand where is the initial position, when the rope is vertical or when the guy is holding on to it?
I have attached the picture , can someone label on the picture or explain it please?

My understanding is, the middle is 8.2m , the right rope is 6.3m, and the left rope is 6.94m
but then the calculation doesn't really make sense.
since if the middle is 8.2, to find l, u should do cos 39.8 = 8.2 / l, and l will be 10.67 instead.

 

[TSny]

The length of the rope is $l$ = 8.2 m. The length of the rope does not change.

Examine some of the triangles shown in the attached figure.

 

[cmkc109]

very helpful, thank you very much.