Question

jaxon is flying a kite, holding his hands a distance of 3.25 feet above the ground and letting all the kites string out. he measures the angle of elevation from his hand to the kite to be 23°. if the string from the kite to his hand is 130 feet long, how many feet is the kite above the ground? round your answer to the nearest hundredth of a foot if necessary.

Explanation:

Step1: Use the sine function

The sine of an angle in a right - triangle is given by \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). Here, \(\theta = 23^{\circ}\) and the hypotenuse (length of the string) is \(130\) feet. Let the vertical distance from the hand to the kite be \(x\). So, \(\sin(23^{\circ})=\frac{x}{130}\).

Step2: Solve for \(x\)

Multiply both sides of the equation \(\sin(23^{\circ})=\frac{x}{130}\) by \(130\). We know that \(\sin(23^{\circ})\approx0.3907\). Then \(x = 130\times\sin(23^{\circ})\approx130\times0.3907 = 50.791\) feet.

Step3: Find the height of the kite above the ground

The height of the kite above the ground \(h\) is the sum of the height of the hand above the ground (\(3.25\) feet) and \(x\). So, \(h=x + 3.25\). Substitute \(x\approx50.791\) into the equation: \(h\approx50.791+3.25=54.041\approx54.04\) feet.

Answer:

The kite is approximately \(54.04\) feet above the ground.