Question

miguel is flying a kite, holding his hands a distance of 3.5 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 35°. if the string from the kite to his hand is 150 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: Find the height from hand level to kite

We use the sine function. In a right - triangle formed by the string (hypotenuse \(c = 150\) feet), the vertical side (height from hand level to kite \(h_1\)) and the horizontal side. The angle of elevation \(\theta=25^{\circ}\).
By the formula \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), so \(h_1 = 150\times\sin(25^{\circ})\).
We know that \(\sin(25^{\circ})\approx0.4226\), then \(h_1=150\times0.4226 = 63.39\) feet.

Step2: Find the height of the kite above the ground

The height of Miguel's hands above the ground \(h_2 = 3.5\) feet.
The total height \(H\) of the kite above the ground is \(H=h_1 + h_2\).
Substitute \(h_1 = 63.39\) and \(h_2 = 3.5\) into the formula: \(H=63.39+3.5=66.89\approx66.9\) feet.

Answer:

\(66.9\) feet