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.

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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.

Physics
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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
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?

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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.\[\cos \theta=\frac{adjacent}{hypotenuse}\]|dw:1443486284490:dw|
Looking at the triangle on the right, you know L and \(\theta_0\), so just plug in the numbers.\[\cos \theta_0=\frac{ h }{ L }\]\[\cos 39°=\frac{ h }{ 7.8 }\]\[h=7.8\cos39°\]
now the triangle on the left. for this one the adjacent side is \(h+d\), and the only thing you don't know is Θ \[\cos \theta = \frac{ h+d }{ L }\] \[\cos \theta = \frac{ 7.8\cos39°+0.76 }{ 7.8 }\] \[\theta=\cos^{-1} \left( \frac{ 7.8\cos39°+0.76 }{ 7.8 } \right)\]

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