To solve these problems, we can use trigonometry and the information provided in the problem.
Given information:
- Angle of the kite with the ground: 50°
- Angle of elevation from the two friends to the kite: 66° and 35°
- Distance of one friend from Carol: 11 m
First, we will calculate how high the kite is above the ground, how long the string is, and how far the other friend is from Carol.
Step 1: Find the height of the kite above the ground (h).
Let \( D \) be the horizontal distance from Carol to the point directly below the kite, and \( L \) be the length of the kite string.
Using the angle formed by the string with the ground (50°):
- The height of the kite \( h \) can be expressed as: \[ h = L \sin(50°) \]
Using the angle of elevation from Friend 1 (66°) and the distance from Carol (11 m): \[ h = 11 \tan(66°) \]
Now we can calculate this height:
- Calculate \( h \): \[ h = 11 \tan(66°) \approx 11 \times 2.244 = 24.684 \approx 25 , \text{m} \]
Conclusion for a):
a) The height of the kite above the ground is approximately 25 m.
Step 2: Find the length of the string (L).
We already know \( h = L \sin(50°) \). From our previous calculation, \( h \approx 25 , m \).
Now we can calculate \( L \): \[ L = \frac{h}{\sin(50°)} \] \[ L = \frac{25}{\sin(50°)} \] \[ \sin(50°) \approx 0.766 \] So, \[ L \approx \frac{25}{0.766} ≈ 32.6 \approx 33 , \text{m} \]
Conclusion for b):
b) The length of the string is approximately 33 m.
Step 3: Find the distance of the other friend from Carol.
Using the angle of elevation from Friend 2 (35°): Let \( d \) be the distance from Friend 2 to Carol.
\[ h = d \tan(35°) \] Since we already know \( h \approx 25 , m \): \[ 25 = d \tan(35°) \] Thus, \[ d = \frac{25}{\tan(35°)} \] Where \( \tan(35°) \approx 0.700 \): \[ d \approx \frac{25}{0.700} \approx 35.71 \approx 36 , m \]
Conclusion for c):
c) The distance of the other friend from Carol is approximately 36 m.
To summarize:
- Height of the kite: ≈ 25 m
- Length of the string: ≈ 33 m
- Distance of the other friend: ≈ 36 m
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