## anonymous one year ago 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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1. anonymous

2. anonymous

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

3. anonymous

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?

4. anonymous

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|

5. anonymous

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°$

6. anonymous

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 $\cos \theta = \frac{ h+d }{ L }$ $\cos \theta = \frac{ 7.8\cos39°+0.76 }{ 7.8 }$ $\theta=\cos^{-1} \left( \frac{ 7.8\cos39°+0.76 }{ 7.8 } \right)$