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Grayson is flying a kite, holding his hands a distance of 2.75 feet above the ground and letting all the kite’s string play out. He measures the angle of elevation from his hand to the kite to be 26
∘
∘
. If the string from the kite to his hand is 145 feet long, how many feet is the kite above the ground? Round your answer to the nearest hundredth of a foot if necessary.

Respuesta :

The position of the kite with the point directly beneath the kite at the same

level with the hand and the hand for a right triangle.

  • The height of the kite above the is approximately 66.314 feet.

Reasons:

The height of his hands above the ground, h = 2.75 feet

Angle of elevation of the string (above the horizontal), θ = 26°

Length of the string, l = 145 feet

Required:

The height of the kite above the ground.

Solution:

The height of the kite above the ground is given by trigonometric ratios as follows;

[tex]\displaystyle Height \ of \ kite, \, k_h = \mathbf{h + l \times sin(\theta)}[/tex]

Therefore;

[tex]\displaystyle Height \ of \ kite, \, k_h = 2.75 + 145 \times sin(26^{\circ}) \approx \mathbf{66.314}[/tex]

The height of the kite above the, [tex]k_h[/tex] ≈ 66.314 feet

Learn more about trigonometric ratios here:

https://brainly.com/question/9085166

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