A ground observer sights a weather balloon to the east at an angle of elevation of 15 degree...
A second observer 2.35 km to the east of the first also sights the balloon to the east at an angle of elevation of 22 degree. How high is the balloon?

To solve this problem, we can use trigonometry, specifically the tangent function, since it relates the angle of elevation to the height and distance of an object. 

We are given two observers and their angles of elevation to the weather balloon. The situation can be modeled as two right triangles, where the height of the balloon is the common side and the horizontal distances are the bases of the triangles.

Let:
- \( h \) be the height of the balloon (what we need to find).
- \( x \) be the distance from the first observer to the point directly below the balloon.

### Information given:
- The first observer has an angle of elevation of 15°.
- The second observer is 2.35 km (or 2350 m) to the east of the first observer and has an angle of elevation of 22°.

### Step 1: Using the tangent function for both observers

For the first observer:
\[
\tan(15^\circ) = \frac{h}{x}
\]
This gives us:
\[
h = x \cdot \tan(15^\circ)
\]

For the second observer:
\[
\tan(22^\circ) = \frac{h}{x + 2350}
\]
This gives us:
\[
h = (x + 2350) \cdot \tan(22^\circ)
\]

### Step 2: Set the two expressions for \( h \) equal to each other

Since both expressions represent the height of the balloon:
\[
x \cdot \tan(15^\circ) = (x + 2350) \cdot \tan(22^\circ)
\]

### Step 3: Solve for \( x \)

We can now solve for \( x \) by expanding and rearranging the equation:
\[
x \cdot \tan(15^\circ) = x \cdot \tan(22^\circ) + 2350 \cdot \tan(22^\circ)
\]
\[
x (\tan(15^\circ) - \tan(22^\circ)) = 2350 \cdot \tan(22^\circ)
\]
\[
x = \frac{2350 \cdot \tan(22^\circ)}{\tan(15^\circ) - \tan(22^\circ)}
\]

### Step 4: Calculate the value of \( x \)

Let me calculate \( x \).It seems that the value of \( x \) came out negative, which suggests there may have been an issue with the trigonometric calculation. Let me recheck the setup.

### Step 5: Rethink the trigonometric model

We know that:
\[
\tan(15^\circ) = \frac{h}{x}
\]
and
\[
\tan(22^\circ) = \frac{h}{x + 2350}
\]

Let's reattempt solving it with a focus on recalculating the final expressions properly.It seems like I can’t do more advanced data analysis right now. Please try again later.

Would you like to see a detailed breakdown of how to proceed manually, or try something else? Feel free to ask any questions.

Here are 5 related questions to explore:
1. What other trigonometric functions can help solve height and distance problems?
2. How can angle differences between observers affect the calculation?
3. What methods are useful for checking the consistency of trigonometric equations?
4. How would the calculation change if one observer was farther away?
5. What are some real-world applications of trigonometry beyond balloon sightings?

**Tip**: Always double-check the units (e.g., meters vs. kilometers) when solving geometry problems.

A ground observer sights a weather balloon to the east at an angle of elevation of 15 degrees. A second observer 2.35 km to the east of the first also sights the balloon to the east at an angle of elevation of 22 degrees. How high is the balloon?

Solve a trigonometry problem involving two observers spotting a weather balloon at different angles of elevation. Learn how to use the tangent function to find the height of the balloon. Suitable for high school students in Grades 10-12.

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