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°

Algebra & Trigonometry with Analytic Geometry
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ISBN:9781133382119
Author:Swokowski
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Chapter11: Topics From Analytic Geometry
Section11.4: Plane Curves And Parametric Equations
Problem 20E
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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
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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