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?11. Carol only on same side is 66°. This is 11 m.
a) 28 m b) 31 m c) 52 m;
Carol only on same side as 66°. Other distance is 11 m.
a) 6 m b) 6 m c) 2 m;
All on same side. Distance to 66° is 11 m.
a) 28 m b) 37 m c) 16 m;
All on same side. Distance to 35° is 11 m.
a) 19 m b) 24 m c) 7 m

Answer 1

To solve the problem, we can use some trigonometric principles involving right triangles to find the height of the kite, the length of the string, and the position of the second friend. We'll break it down as follows:

Given Information:

• The angle of elevation to the kite from one friend at a distance of 11 m from Carol is 66°.
• The angle of elevation to the kite from the other friend is 35°.
• The angle the string makes with the ground is 50°.

a) How high is the kite above the ground?

1. For the friend at 66°:

   • Height of kite \( h_1 \) can be calculated using: \[ h_1 = 11 \tan(66°) \] \[ h_1 \approx 11 \times 2.246 = 24.7 \text{ m} \text{ (approximately 25 m)} \]

2. For the friend at 35°:

   • Let \( d \) be the distance of the second friend from Carol, and we can denote height of kite \( h_2 \) as: \[ h_2 = d \tan(35°) \]

b) How long is the string?

To find the length of the string \( L \):

Using the angle the string makes with the ground (50°): \[ L \cos(50°) = h \] Where \( h \) is the height of the kite.

Using \( h_1 \) (from friend at 66°): \[ L = \frac{h_1}{\cos(50°)} \rightarrow L \approx \frac{25}{0.643} \approx 38.9 \text{ m} \text{ (approximately 39 m)} \]

c) How far is the other friend from Carol?

Using the height found from the friend at 35°: \[ d = \frac{h_2}{\tan(35°)} \]

We can rewrite this using height \( h_1 \): \[ d = \frac{h_1}{\tan(35°)} \approx \frac{25}{0.700} \approx 35.7 \text{ m} \text{ (approximately 36 m)} \]

Summary of Results:

1. Height of the Kite (to the nearest metre):
   • 25 m
2. Length of the String (to the nearest metre):
   • 39 m
3. Distance of the Other Friend from Carol (to the nearest metre):
   • 36 m

These results are based on the calculated heights and distances, and specifics of other configurations (like both friends being on the same side) could yield different results, but these represent one valid scenario based on the given angles.