An aerialist on a high platform holds onto a trapeze attached to a support by an 8.0-m cord. (See the drawing.) Just before he jumps off the platform, the cord makes an angle of 41° with the vertical. He jumps, swings down, then back up, releasing the trapeze at the instant it is 0.75 m below its initial height. Calculate the angle θ that the trapeze cord makes with the vertical at this instant.
Imagine two triangles:
1) ΔABC : point A (where the cord is attached ), point B (where the aerialist holds the trapeze at initial position), point C (the intersection of the vertical line from A and horizontal line from B),
2) ΔADM: point M (where the aerialist holds the trapeze at its left position), point D (the intersection of the vertical line from A and horizontal line from M)
AC = AB•cosθₒ =L•cosθₒ,
AD =AM•cosθ = L•cosθ,
CD = d.
AD =AC +CD,
L•cosθ = L•cosθₒ +d.
cosθ =( L•cosθₒ +d)/L =
=(8•cos41º+0.75)/8 = 0.848.
arcos 0.848 = 32º.
arccos 0.848 = 32º.
To solve this problem, we can break it down into two parts: the initial position and the final position.
First, let's find the initial angle the trapeze cord makes with the vertical. We are given that the cord makes an angle of 41° with the vertical just before the aerialist jumps off the platform.
Next, let's find the final angle the cord makes with the vertical when the trapeze is 0.75 m below its initial height.
Using the information provided, we can draw a right triangle to represent these positions. Let's label the initial angle as θ1 and the final angle as θ2.
Now, we can use trigonometric ratios to solve for these angles.
In the initial position:
sin(θ1) = opposite / hypotenuse
sin(41°) = opposite / 8.0 m
opposite = 8.0 m * sin(41°)
opposite ≈ 5.28 m
In the final position:
sin(θ2) = opposite / hypotenuse
sin(θ2) = 0.75 m / 8.0 m
θ2 = arcsin(0.75 m / 8.0 m)
θ2 ≈ 5.84°
Therefore, the angle θ that the trapeze cord makes with the vertical at this instant is approximately 5.84°.
To calculate the angle θ that the trapeze cord makes with the vertical at the instant the trapeze is 0.75 m below its initial height, we can use trigonometry.
Let's break down the problem into two parts: the initial position and the final position of the trapeze.
1. Initial position:
Before the aerialist jumps off the platform, the cord makes an angle of 41° with the vertical. From this information, we can draw a right-angled triangle, with the cord being the hypotenuse, the vertical line as one side, and the horizontal line as the other side.
Using trigonometry, we can use the sine function to find the length of the vertical side.
sin(41°) = opposite/hypotenuse
sin(41°) = vertical length/8.0 m
Now solve for the vertical length:
vertical length = 8.0 m * sin(41°)
2. Final position:
When the trapeze is 0.75 m below its initial height, we need to find the angle θ that the trapeze cord makes with the vertical.
Once again, let's consider a right-angled triangle. One side will be the vertical line, the other side will be the horizontal distance (0.75 m), and the hypotenuse will be the cord length (8.0 m).
Using trigonometry, we can use the inverse sine function to find the angle θ.
sin^(-1)(opposite/hypotenuse) = θ
sin^(-1)(0.75 m/8.0 m) = θ
Now, calculate the value of θ.
By solving the equation, we find that θ is equal to sin^(-1)(0.09375) which is approximately 5.4°.
Therefore, the angle θ that the trapeze cord makes with the vertical at the instant it is 0.75 m below its initial height is approximately 5.4°.