To find the horizontal distance \( x \) that the plane has flown when it is at an altitude of 250 feet and taking off at a 15° angle, we can use the trigonometric relationship of the sine function.
In this context, we can describe the situation as follows:
- The altitude of the plane (opposite side) is 250 feet.
- The angle of takeoff is 15°.
- The horizontal distance (adjacent side) to be found is \( x \).
Using the sine function:
\[ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \]
In our case, we can also relate the horizontal distance \( x \) to these sides with the tangent function:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
Thus,
\[ \tan(15°) = \frac{250}{x} \]
Now we can rearrange this to solve for \( x \):
\[ x = \frac{250}{\tan(15°)} \]
Next, we need to calculate \( \tan(15°) \):
\[ \tan(15°) \approx 0.2679 \]
Now, substituting this value into the equation for \( x \):
\[ x = \frac{250}{0.2679} \approx 933.33 \]
Rounding this to the nearest whole number:
\[ x \approx 933 \]
So, the horizontal distance \( x \) the plane has flown is approximately 933 feet.
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