1. The problem statement, all variables and given/known data 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. 2. Relevant equations cos(theta) = A/H 3. 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 dont 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 doesnt 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.