An airplane is flying at an altitude of 10,000 feet. At a certain point along its flight path, it starts descending with a rate of descent of 500 feet per minute. How many degrees does the airplane's descent make with the horizontal after 15 minutes?
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Let's call the angle the airplane's descent makes with the horizontal $\theta$. We can use trigonometry to find $\theta$. We know that the rate of descent is 500 feet per minute and the time is 15 minutes, so the airplane descends $500 \cdot 15 = 7500$ feet. We can use this information to find the horizontal distance the airplane travels during this time.
We can set up a right triangle with the horizontal distance as the adjacent side, the vertical distance as the opposite side, and the hypotenuse as the distance the airplane travels. To find the horizontal distance, we can use the tangent function, which is defined as $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$. In this case, $\frac{\text{opposite}}{\text{adjacent}} = \frac{7500}{\text{horizontal distance}}$.
Rearranging this equation, we find that the horizontal distance is $\text{horizontal distance} = \frac{7500}{\tan \theta}$.
We can use the Pythagorean theorem to find the hypotenuse, which is the distance the airplane travels. The Pythagorean theorem is defined as $a^2 + b^2 = c^2$, where $a$ and $b$ are the legs of the right triangle and $c$ is the hypotenuse. In this case, $a$ is the horizontal distance and $b$ is the vertical distance.
Plugging in the values we know, we have $(\frac{7500}{\tan \theta})^2 + 7500^2 = \text{distance}^2$. Solving for the distance, we find that the distance is $\text{distance} = \sqrt{(\frac{7500}{\tan \theta})^2 + 7500^2}$.
Now, we can use this information to find $\theta$. We know that $\tan \theta = \frac{7500}{\text{horizontal distance}}$, so $\theta = \arctan (\frac{7500}{\text{horizontal distance}})$.
Plugging in the values we know, we have $\theta = \arctan (\frac{7500}{\sqrt{(\frac{7500}{\tan \theta})^2 + 7500^2}})$.
Using a calculator, we find that $\theta \approx \arctan (\frac{7500}{8821.08}) \approx \arctan (0.85014) \approx 40.9$ degrees.
Therefore, the airplane's descent makes an angle of approximately 40.9 degrees with the horizontal after 15 minutes.
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