Calculating Balloon Height Using Angles of Elevation

This problem asks us to find the height of a balloon based on the angles of elevation measured from two consecutive kilometer-stones along a straight road. We need to use trigonometry to solve this.

Understanding the Setup

Let's visualize the scenario:

• A balloon is at a certain height above the ground.
• There's a straight road.
• Two observation points are located at consecutive kilometer-stones on this road. The distance between these two stones is 1 km.
• The angles of elevation from these stones to the balloon are given as $30^\circ$ and $60^\circ$.

Setting up the Trigonometric Equations

Let $h$ be the height of the balloon above the ground. Let the point directly below the balloon on the ground be $P$. Let the two kilometer-stones be $A$ and $B$, such that $A$ is closer to $P$ than $B$. The distance between $A$ and $B$ is $1$ km.

Let the distance from the closer stone ($A$) to point $P$ be $x$ km.

The distance from the farther stone ($B$) to point $P$ is then $(x + 1)$ km.

We can form two right-angled triangles:

1. Triangle APB' (where B' is the balloon):
   • Angle of elevation at A: $30^\circ$
   • Opposite side (height): $h$
   • Adjacent side (distance): $x$
   • Using the tangent function: $\tan(30^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{h}{x}$
   • We know $\tan(30^\circ) = \frac{1}{\sqrt{3}}$. So, $\frac{1}{\sqrt{3}} = \frac{h}{x}$.
   • Rearranging this, we get $x = h\sqrt{3}$.
2. Triangle B'PB (where B' is the balloon):
   • Angle of elevation at B: $60^\circ$
   • Opposite side (height): $h$
   • Adjacent side (distance): $x + 1$
   • Using the tangent function: $\tan(60^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{h}{x+1}$
   • We know $\tan(60^\circ) = \sqrt{3}$. So, $\sqrt{3} = \frac{h}{x+1}$.
   • Rearranging this, we get $x+1 = \frac{h}{\sqrt{3}}$.

Note: The angles seem swapped in the standard convention where the closer point has a larger angle of elevation. However, we will proceed with the values as given in the question text. Let's assume the angle from the closer stone is $60^\circ$ and the farther is $30^\circ$, which is physically more intuitive.

Revised Setup (Intuitive Angles)

Let's assume the angle of elevation from the closer stone ($A$) is $60^\circ$ and from the farther stone ($B$) is $30^\circ$. The distance $AB = 1$ km.

Let $h$ be the height and $x$ be the distance $AP$. The distance $BP = x+1$.

1. From stone A (closer, angle $60^\circ$): $\tan(60^\circ) = \frac{h}{x} \implies \sqrt{3} = \frac{h}{x} \implies x = \frac{h}{\sqrt{3}}$
2. From stone B (farther, angle $30^\circ$): $\tan(30^\circ) = \frac{h}{x+1} \implies \frac{1}{\sqrt{3}} = \frac{h}{x+1} \implies x+1 = \frac{h}{1/\sqrt{3}} = h\sqrt{3}$

Solving for the Height ($h$)

Now we substitute the expression for $x$ from the first equation into the second equation:

$$ \left( \frac{h}{\sqrt{3}} \right) + 1 = h\sqrt{3} $$

To solve for $h$, we first gather the terms involving $h$ on one side:

$$ 1 = h\sqrt{3} - \frac{h}{\sqrt{3}} $$

Factor out $h$:

$$ 1 = h \left( \sqrt{3} - \frac{1}{\sqrt{3}} \right) $$

Find a common denominator for the terms in the parenthesis:

$$ 1 = h \left( \frac{3}{\sqrt{3}} - \frac{1}{\sqrt{3}} \right) $$

$$ 1 = h \left( \frac{3-1}{\sqrt{3}} \right) $$

$$ 1 = h \left( \frac{2}{\sqrt{3}} \right) $$

Now, isolate $h$:

$$ h = \frac{1}{\frac{2}{\sqrt{3}}} $$

$$ h = \frac{\sqrt{3}}{2} \text{ km} $$

Conclusion

The height of the balloon above the ground is $\frac{\sqrt{3}}{2}$ km.