A balloon is moving up in air vertically above a point A on the ground. When it is at a height h1, a

Q: Why did we use $45^\circ$ and $30^\circ$ instead of $45^\circ$ and $60^\circ$?

The problem gives angles with respect to the vertical. For standard right-triangle trigonometry (using the ground as the base), we need the angles of elevation, which are measured with respect to the horizontal. Since the vertical and horizontal are perpendicular, the angle of elevation is $90^\circ$ minus the angle with the vertical.

Q: How did $0.5774 \times 3.464$ become exactly 2?

It helps to recognize the underlying surds. $\tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.5774$. The number $3.464$ is $2 \times 1.732$, which is $2\sqrt{3}$. When you multiply them: $\frac{1}{\sqrt{3}} \times 2\sqrt{3}$, the $\sqrt{3}$ terms cancel out, leaving exactly 2.

Q: Why is the total base distance $3.464d$?

The girl was initially at a distance $d$ from the launch point. The problem states she moves *further* by a distance of $2.464d$. Therefore, her new total distance from the launch point is the sum of these two distances: $d + 2.464d = 3.464d$.

Q: Can I solve this without knowing the value of $\sqrt{3}$?

Yes! The problem explicitly provides $\tan 30^\circ = 0.5774$. You can simply multiply the decimals $0.5774 \times 3.464$ on paper or in your head, and you will get $1.9999...$, which safely rounds to $2$ for the context of this problem.

Q: What if the balloon didn't move perfectly vertically?

If the balloon had horizontal motion (like being blown by wind), we would need more information, such as its horizontal displacement, to form the correct right-angled triangles. The assumption of vertical motion is critical here.

Question

A balloon is moving up in air vertically above a point A on the ground. When it is at a height $h_1$, a girl standing at a distance $d$ (point B) from A (see figure) sees it at an angle $45^\circ$ with respect to the vertical. When the balloon climbs up a further height $h_2$, it is seen at an angle $60^\circ$ with respect to the vertical if the girl moves further by a distance $2.464 d$ (point C). Then, the height $h_2$ is
(Given, $	an 30^\circ = 0.5774$)

Options:

(A) $1.464 d$

(B) $d$

(C) $0.464 d$

(D) $0.732 d$

Sigma Percentile · Accepted Answer

### The First Sighting: Establishing the Baseline

Imagine you are standing on a flat ground, watching a hot air balloon rise straight up into the sky. This problem is a classic application of trigonometry to real-world observations.

Let's break down the first scenario. The balloon is at a height $h_1$ directly above point A. A girl is standing at point B, which is a horizontal distance $d$ away from A. The problem states she sees the balloon at an angle of $45^\circ$ *with respect to the vertical*.

Here is the crucial geometric insight: if the line of sight makes a $45^\circ$ angle with the vertical, it must also make a $90^\circ - 45^\circ = 45^\circ$ angle with the horizontal ground. This is our **angle of elevation**.

Looking at the right-angled triangle $\Delta ABD$, we can apply the tangent function:

$$	an 45^\circ = \frac{	ext{Perpendicular}}{	ext{Base}} = \frac{h_1}{d}$$

Since we know that $	an 45^\circ = 1$, this simplifies beautifully to:

$$1 = \frac{h_1}{d} \implies h_1 = d$$

This gives us a solid baseline: the initial height of the balloon is exactly equal to the girl's initial distance from the launch point.

### The Second Sighting: A Wider Perspective

Now, the situation evolves. The balloon climbs higher by an additional height $h_2$, reaching a total height of $h_1 + h_2$. Simultaneously, the girl walks further away from the launch point by a distance of $2.464d$, arriving at a new point C.

Her total distance from the launch point A is now:

$$AC = d + 2.464d = 3.464d$$

From this new vantage point, she looks up again. This time, the line of sight makes a $60^\circ$ angle with the vertical. Using the same logic as before, the new angle of elevation is $90^\circ - 60^\circ = 30^\circ$.

We now focus on the larger right-angled triangle $\Delta ACE$. Applying the tangent function again:

$$	an 30^\circ = \frac{	ext{Total Height}}{	ext{Total Base}} = \frac{h_1 + h_2}{3.464d}$$

### The Mathematical Magic: Bringing It Together

We are given the value $	an 30^\circ = 0.5774$. Let's substitute this into our equation, and also replace $h_1$ with $d$ (which we found in the first step):

$$0.5774 = \frac{d + h_2}{3.464d}$$

To solve for $h_2$, we cross-multiply:

$$d + h_2 = 0.5774 	imes 3.464d$$

Here is where the numbers work out like magic. If you recognize that $0.5774$ is approximately $\frac{1}{\sqrt{3}}$ and $3.464$ is exactly $2 	imes 1.732$ (which is $2\sqrt{3}$), their product is exactly $2$. Even if you just multiply the decimals, $0.5774 	imes 3.464 = 1.9999... \approx 2$.

So, the equation simplifies to:

$$d + h_2 = 2d$$

Subtracting $d$ from both sides, we arrive at our final, elegant conclusion:

**$$h_2 = d$$**

The additional height the balloon climbed is exactly equal to the initial distance $d$.

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Visualized Solution

$Visualizing the Setup$

$Analyzing Δ A B D$

$First Equation$

$The Second Observation$

$Analyzing Δ A C E$

$Second Equation Setup$

$Substitution$

$Simplification$

$Final Answer$

The Sigma Insight: Motion in a Straight Line

The First Sighting

The Second Sighting

The Mathematical Magic

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing ΔABD

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Second Observation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing ΔACE

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Second Equation Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplification

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer