Question

1. brian’s kite is flying above a field at the end of 65 m of string. if the angle of elevation to the kite measures 70°, and brian is holding the kite 1.2 m off the ground. how high above the ground is the kite flying?

Explanation:

Step1: Use sine function

Let \(h\) be the height from Brian's hand level to the kite. We know that \(\sin\theta=\frac{opposite}{hypotenuse}\). Here, \(\theta = 70^{\circ}\) and hypotenuse \(= 65\) m. So, \(\sin70^{\circ}=\frac{h}{65}\).

Step2: Solve for \(h\)

\(h = 65\times\sin70^{\circ}\). Since \(\sin70^{\circ}\approx0.9397\), then \(h\approx65\times0.9397 = 61.0805\) m.

Step3: Calculate total height

The total height of the kite above the ground is \(H=h + 1.2\) m. Substitute \(h\approx61.0805\) m, so \(H\approx61.0805+1.2=62.2805\approx62.3\) m.

Answer:

The kite is flying approximately \(62.3\) m above the ground.