ubrey is flying a kite, holding her hands a distance of 2.5 feet above the ground and letting all the kite's string play out. She measures the angle of elevation from her hand to the kite to be \( 25^{\circ} \). If the string from the kite to her hand is 150 feet long, how many feet is the kite above the ground? Round your answer to the nearest hundredth of a foot if necessary.

To find out how high the kite is above the ground, we can use some trigonometry! First, we determine the vertical height (h) of the kite above where she is holding the string using the sine function. The angle of elevation to the kite is \( 25^{\circ} \), and the length of the kite string is 150 feet. The formula we use is: \[ h = l \cdot \sin(\theta) \] where \( l = 150 \) feet is the length of the string and \( \theta = 25^{\circ} \). Calculating: \[ h = 150 \cdot \sin(25^{\circ}) \] Using a calculator, we find that \( \sin(25^{\circ}) \approx 0.4226 \). So: \[ h \approx 150 \cdot 0.4226 \approx 63.39 \text{ feet} \] Now, to get the total height of the kite above the ground, we need to add the height of her hands above the ground: \[ \text{Total Height} = h + 2.5 \text{ feet} \approx 63.39 + 2.5 = 65.89 \text{ feet} \] Thus, the kite is approximately \( 65.89 \) feet above the ground.