Question

jaxon is flying a kite, holding his hands a distance of 3.25 feet above the ground and letting all the kites string play out. he measures the angle of elevation from his hand to the kite to be 24°. if the string from the kite to his hand is 105 feet long, how many feet is the kite above the ground? round your answer to the nearest tenth of a foot if necessary.

Explanation:

Step1: Identify vertical component height

We use the sine function to find the vertical distance from Jaxon's hand to the kite, since $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$.
$\text{Vertical height from hand} = 105 \times \sin(24^\circ)$
Calculate $\sin(24^\circ) \approx 0.4067$, so $105 \times 0.4067 \approx 42.7035$ feet.

Step2: Add hand height to ground

Add the height of Jaxon's hands above the ground to the vertical height from his hand to the kite.
$\text{Total height above ground} = 42.7035 + 3.25$

Answer:

45.9 feet