Question

An aerialist on a high platform holds on to a trapeze attached to a support by a cord of length L = 7.8 m. (See the drawing below.) Just before he jumps off the platform, the cord makes an angle θ0 = 39° with the vertical. He jumps, swings down, then back up, releasing the trapeze at the instant it is a distance d = 0.76 m below its initial height. Calculate the angle θ that the trapeze cord makes with the vertical at this instant.

Answer 1

Extend the dashed vertical line between the angles to complete two right triangles, each with the cord as the hypotenuse.

Let h be the initial height. Solve for h.
\[\cos \theta_0=\frac{ h }{ L }\]
Then since the cord length doesn't change
\[\cos \theta = \frac{ h+d }{ L }\]Solve for Θ

Answer 2

ive never taken trig or precalc and need it as a prerequisite for physics but my guidance counselor stuck me in it anyway so now im really struggling with the homework :( do you think you can show me step by step how to solve the problem?

Answer 3

It's trig, but mostly what's taught in geometry. review SOHCAHTOA and that should cover 90% of the trig you need. Specifically for this problem you need to know the cosine ratio.\[\cos \theta=\frac{adjacent}{hypotenuse}\]