A balloon is connected to a meteorological station by a cable 200 meters long inclined at a 60 degree angle with the ground. Find the height of the A balloon from the ground. 200 m 60°
A balloon is connected to a meteorological station by a cable 200 meters long inclined at a 60 degree angle with the ground. Find the height of the A balloon from the ground. 200 m 60°
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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![### Hot Air Balloon Height Calculation
A hot air balloon is tethered to a meteorological station by a cable that is 200 meters in length. This cable is inclined at a 60-degree angle with respect to the ground. Our goal is to determine the height of the balloon from the ground.
#### Diagram Explanation
In the diagram provided:
- Point **A** represents the position of the hot air balloon.
- Point **C** is a fixed point on the ground directly below point **A**.
- Point **B** is the point on the ground right below where the cable is anchored to the meteorological station.
- The length of the cable, which is the hypotenuse of the right triangle formed, is 200 meters.
- The angle formed between the ground and the cable (angle BAC) is 60 degrees.
- Side **AB** (unknown height x) is what we need to determine.
#### Mathematical Approach
The situation forms a right-angled triangle \( \triangle ABC \) with:
- Hypotenuse \( \overline{AC} = 200 \) meters
- Adjacent side \( \overline{BC} \) and opposite side \( \overline{AB} = x \)
Using trigonometry, specifically the sine function:
\[
\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}
\]
Here, \( \theta = 60^\circ \), the opposite side is \( \overline{AB} \) (height x), and the hypotenuse is \( \overline{AC} \).
\[
\sin(60^\circ) = \frac{x}{200}
\]
From trigonometric values, we know that:
\[
\sin(60^\circ) = \frac{\sqrt{3}}{2}
\]
Thus:
\[
\frac{\sqrt{3}}{2} = \frac{x}{200}
\]
Solving for \( x \):
\[
x = 200 \times \frac{\sqrt{3}}{2}
\]
\[
x = 100\sqrt{3}
\]
Taking \( \sqrt{3} \approx 1.732 \):
\[
x \approx 100 \times 1.732 = 173.2 \text{ meters}
\]
So, the height of the balloon from the ground is approximately **173.2 meters](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc476fe49-bd29-4169-8079-6a3850fe1835%2F64fe71b0-8a52-4825-bf02-381623625510%2Fd67msjg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Hot Air Balloon Height Calculation
A hot air balloon is tethered to a meteorological station by a cable that is 200 meters in length. This cable is inclined at a 60-degree angle with respect to the ground. Our goal is to determine the height of the balloon from the ground.
#### Diagram Explanation
In the diagram provided:
- Point **A** represents the position of the hot air balloon.
- Point **C** is a fixed point on the ground directly below point **A**.
- Point **B** is the point on the ground right below where the cable is anchored to the meteorological station.
- The length of the cable, which is the hypotenuse of the right triangle formed, is 200 meters.
- The angle formed between the ground and the cable (angle BAC) is 60 degrees.
- Side **AB** (unknown height x) is what we need to determine.
#### Mathematical Approach
The situation forms a right-angled triangle \( \triangle ABC \) with:
- Hypotenuse \( \overline{AC} = 200 \) meters
- Adjacent side \( \overline{BC} \) and opposite side \( \overline{AB} = x \)
Using trigonometry, specifically the sine function:
\[
\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}
\]
Here, \( \theta = 60^\circ \), the opposite side is \( \overline{AB} \) (height x), and the hypotenuse is \( \overline{AC} \).
\[
\sin(60^\circ) = \frac{x}{200}
\]
From trigonometric values, we know that:
\[
\sin(60^\circ) = \frac{\sqrt{3}}{2}
\]
Thus:
\[
\frac{\sqrt{3}}{2} = \frac{x}{200}
\]
Solving for \( x \):
\[
x = 200 \times \frac{\sqrt{3}}{2}
\]
\[
x = 100\sqrt{3}
\]
Taking \( \sqrt{3} \approx 1.732 \):
\[
x \approx 100 \times 1.732 = 173.2 \text{ meters}
\]
So, the height of the balloon from the ground is approximately **173.2 meters
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