To solve the problem, we can use trigonometry.
Part 1: Calculate the height of the flagpole
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For Jeffery: The angle of elevation from Jeffery to the flag is 45 degrees. We can use the tangent function, which relates the angle of elevation to the opposite side (height of the flagpole) and the adjacent side (distance from Jeffery to the flagpole).
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
For Jeffery:
- Angle \(\theta = 45^\circ\)
- Adjacent side = 10 m (distance from Jeffery to the base of the flagpole)
- Let \(h\) be the height of the flagpole.
\[ \tan(45^\circ) = \frac{h}{10} \]
Since \(\tan(45^\circ) = 1\):
\[ 1 = \frac{h}{10} \implies h = 10 \text{ m} \]
So, the height of the flagpole is 10 meters.
Part 2: How far are Jeffery and Nicole standing from each other
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For Nicole: The angle of elevation from Nicole to the flag is 30 degrees. We can use the tangent function again.
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
For Nicole:
- Angle \(\theta = 30^\circ\)
- Height \(h = 10 \text{ m}\)
- Let \(d\) be the distance from Nicole to the base of the flagpole.
\[ \tan(30^\circ) = \frac{10}{d} \]
Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\):
\[ \frac{1}{\sqrt{3}} = \frac{10}{d} \]
Rearranging gives:
\[ d = 10 \sqrt{3} \text{ m} \]
Now, we calculate \(10 \sqrt{3}\):
\[ d \approx 10 \times 1.732 \approx 17.32 \text{ m} \]
Distance between Jeffery and Nicole:
Since Jeffery is standing 10 m from the flagpole and Nicole is standing approximately 17.32 m from the flagpole:
\[ \text{Distance between Jeffery and Nicole} = d + 10 = 17.32 + 10 \approx 27.32 \text{ m} \]
In summary:
- The height of the flagpole is approximately 10 meters.
- Jeffery and Nicole are standing approximately 27.32 meters apart.
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