Answer: the height of the kite is 106.065 ft
Step-by-step explanation:
The length of the kite represents the hypotenuse of the right angle triangle. The height of the kite represents the opposite side of the right angle triangle.
To determine x, the height of the kite, we would apply the sine trigonometric ratio which is expressed as
Sin θ = opposite side/hypotenuse
Therefore,
Sin 45 = x/150
Cross multiplying, it becomes
x = 150Sin45 = 150 × 0.7071
x = 106.065 ft
Using proportions and the information given, it is found that:
- The class width is of 14.375.
- The lower class limits are: {19, 33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625}.
- The upper class limits are: {33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625, 134}.
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- Minimum value is 19.
- Maximum value is of 134.
- There are 8 classes.
- The classes are all of equal width, thus the width is of:
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The intervals will be of:
19 - 33.375
33.375 - 47.750
47.750 - 62.125
62.125 - 76.500
76.500 - 90.875
90.875 - 105.250
105.250 - 119.625
119.625 - 134.
- The lower class limits are: {19, 33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625}.
- The upper class limits are: {33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625, 134}.
A similar problem is given at brainly.com/question/16631975
The two horizontal lines are parallel ( This is given by the red triangles)
Because the lines are parallel, the two angles 1 & 2 would be the same.
Set the two equations for the angle to equal each other and solve:
8y-6 = 7y
Add 6 to each side:
8y = 7y+6
Subtract 7y from each side:
y = 6
Now you have the value for y, solve for angle 2 by replacing y with 6:
Angle 2 = 7y = 7(6) = 42 degrees.
Choice given:
<span>33.6°
39.8°
50.2°
56.4°
I drew the figure.
I got 12 ft as the hypotenuse, 10 ft as the opposite.
Sin</span>Θ = opposite / hypotenuse
SinΘ = 10/12
SinΘ = 0.83
I used the calculator to get the value of each angle using the sine function
sin(33.6°) = 0.55
sin(39.8°) = 0.64
sin(50.2°) = 0.77
sin(56.4°) = 0.83
The angle where the wire meets the ground is approximately 56.4°
