1. **State the problem:** We have a kite flying such that from point A on the ground, the string makes a 75° angle of elevation, and from point B, which is 240 ft horizontally from A, the angle of elevation is 45°. We want to find the length of the kite string from point A to the kite. 2. **Set up the scenario:** Let the kite be at point K. Points A and B are on the ground, with AB = 240 ft. The angles of elevation at A and B are $75^\circ$ and $45^\circ$ respectively. 3. **Use trigonometry:** Let the horizontal distance from A to the kite's vertical projection on the ground be $x$ ft. Then the horizontal distance from B to that projection is $240 - x$ ft. 4. **Express the height of the kite:** From point A, height $h = x \tan 75^\circ$. From point B, height $h = (240 - x) \tan 45^\circ$. 5. **Set the heights equal:** $$ x \tan 75^\circ = (240 - x) \tan 45^\circ $$ 6. **Substitute values:** $$ \tan 75^\circ \approx 3.732, \quad \tan 45^\circ = 1 $$ So, $$ 3.732x = 240 - x $$ 7. **Solve for $x$:** $$ 3.732x + x = 240 $$ $$ \cancel{3.732x} + \cancel{x} = 240 $$ $$ 4.732x = 240 $$ $$ x = \frac{240}{4.732} \approx 50.7 $$ 8. **Find the height $h$:** $$ h = x \tan 75^\circ = 50.7 \times 3.732 \approx 189.3\text{ ft} $$ 9. **Find the length of the string $AK$:** $$ AK = \frac{h}{\sin 75^\circ} \quad \text{or} \quad AK = \frac{x}{\cos 75^\circ} $$ Using $AK = \frac{x}{\cos 75^\circ}$: $$ \cos 75^\circ \approx 0.2588 $$ $$ AK = \frac{50.7}{0.2588} \approx 195.8\text{ ft} $$ **Final answer:** The length of the kite string is approximately **196 feet**.