Carol is flying a kite on level ground, and the string forms an angle of 50

Question

with the ground. Two friends standing some distance from Carol see the kite
at angles of elevation of 66 and 35, respectively. One friend is 11 m from
Carol. For each question below, state all possible answers to the nearest metre.
a) How high is the kite above the ground?
b) How long is the string?
c) How far is the other friend from Carol?

Answer 1

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:

1. Height of the kite: ≈ 25 m
2. Length of the string: ≈ 33 m
3. Distance of the other friend: ≈ 36 m