An aerialist on a high platform holds onto a trapeze attached to a support by an 8.0 m cord. Just before he jumps off the platform, the cord makes an angle of 41° with the vertical. He swings to the other side, releasing the trapeze at the instant it is 0.75 m below its initial height. Calculate the angle (in degrees) that the trapeze cord makes with the vertical at this instant.

Using the law of sines, the new angle with the vertical made by the trapeze cord when the acrobat releases it 0.75 m lower than its initial height is about 38.38 degrees.

Explanation:

The subject of this question is Physics, specifically trigonometry and kinematics. The grade is High School.

To answer the question, we use the law of sines in trigonometry that states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is constant for all three sides of the triangle.

We know at first the trapeze cord makes an angle of 41 degrees with the vertical and the length of the cord remains 8.0 m. And as the aerialist jumps off the platform and the trapeze cord adjusted itself, the angle changes such that the height from the initial point is 0.75 m less.

Let's denote the new angle with the vertical as α, so we have sinα / sin(180 - α - 41) = (8.0 - 0.75) / 8.0. Solving this equation will get the α around 38.38 degrees.

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