Question

kaj is flying a kite, holding her hands a distance of 3.5 feet above the ground and letting all the kites string play out. she measures the angle of elevation from her hand to the kite to be 23°. if the string from the kite to her hand is 75 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: Find the vertical height from her hand to the kite

We use the sine - function in a right - triangle. Let the vertical height from her hand to the kite be $h$. The length of the string is the hypotenuse $c = 75$ feet and the angle of elevation $\theta=23^{\circ}$. The formula for sine is $\sin\theta=\frac{opposite}{hypotenuse}$. So, $h = c\times\sin\theta$.
$h = 75\times\sin(23^{\circ})$
Since $\sin(23^{\circ})\approx0.3907$, then $h = 75\times0.3907 = 29.3025$ feet.

Step2: Find the height of the kite above the ground

The height of her hand above the ground is $3.5$ feet. The height of the kite above the ground $H$ is the sum of the height of her hand above the ground and the vertical height from her hand to the kite.
$H=h + 3.5$
Substitute $h = 29.3025$ into the equation: $H=29.3025+3.5=32.8025\approx32.80$ feet.

Answer:

$32.80$ feet