Two vertical poles of heights, 20m and 80m stand a part on a horizontal plane. The height (in meters

Q: Wait, why does the distance $d$ between the poles not matter? Does the intersection point not shift if I move the poles further apart?

That is a brilliant observation! It feels counterintuitive, right? But mathematically, as you increase $d$, the lines become less steep, but they stretch proportionally. The ratio of the heights remains constant because the triangles scale uniformly. The intersection height $h$ is purely a function of the pole heights $H_1$ and $H_2$.

Q: Can I solve this using coordinate geometry by finding the equations of the two lines and solving for their intersection?

Absolutely! You can place the foot of the first pole at $(0,0)$ and the second at $(d,0)$. The lines are $y = \frac{H_1}{d}x$ and $y = -\frac{H_2}{d}(x-d)$. Solving this system will yield the same result. However, the similar triangle approach is much faster and less prone to algebraic errors in an exam setting.

Q: Is this formula $\frac{1}{h} = \frac{1}{H_1} + \frac{1}{H_2}$ valid for any shape of poles?

Great question! This specific derivation relies on the poles being vertical and parallel. If the poles were tilted or not parallel, the geometry would change drastically. Always check that the 'poles' are parallel vertical lines before applying this 'reciprocal sum' shortcut.

Question

Two vertical poles of heights, 20m and 80m stand a part on a horizontal plane. The height (in meters) of the point of intersection of the lines joining the top of each pole to the foot of the other, from this horizontal plane is :

Options:

(A) 12

(B) 15

(C) 16

(D) 18

Sigma Percentile · Accepted Answer

### The Geometry of Elegance: Solving the Pole Intersection Problem

Imagine you are standing on a perfectly flat, horizontal plane. In front of you, two vertical poles rise like sentinels—one is $20	ext{ m}$ tall, and the other is $80	ext{ m}$ tall.

You draw a line from the top of the first pole to the base of the second, and another from the top of the second to the base of the first. They cross. At what height does this intersection occur?

This is a classic JEE problem, and it is a beautiful example of how geometry can simplify a seemingly complex scenario.

### Phase 1: The Illusion of Complexity

When you first look at this problem, your brain might scream, "I need the distance between the poles!" It is a natural instinct. We are trained to look for all the variables.

But here is the secret: the distance $d$ is a red herring. It is a ghost variable.

Whether the poles are $10	ext{ m}$ apart or $100	ext{ m}$ apart, the intersection height $h$ remains unchanged. This is because the geometry is governed by ratios, not absolute lengths.

### Phase 2: The Power of Similar Triangles

Let the height of the intersection point be $h$. Let the distance between the poles be $d$. Let the intersection point be at a distance $x$ from the first pole.

By the property of similar triangles, we can establish the following relationships:

$$
\frac{h}{H_1} = \frac{d-x}{d}
$$

$$
\frac{h}{H_2} = \frac{x}{d}
$$

Do you see the beauty here? We have expressed the height $h$ in terms of the pole heights and the horizontal segments. If we add these two equations, something magical happens:

$$
\frac{h}{H_1} + \frac{h}{H_2} = \frac{d-x}{d} + \frac{x}{d}
$$

$$
\frac{h}{H_1} + \frac{h}{H_2} = \frac{d-x+x}{d} = \frac{d}{d} = 1
$$

### Phase 3: The Reciprocal Sum Rule

We have arrived at the core of the problem: $\frac{h}{H_1} + \frac{h}{H_2} = 1$. If we divide both sides by $h$, we get the famous reciprocal sum rule:

$$
\frac{1}{h} = \frac{1}{H_1} + \frac{1}{H_2}
$$

This is the "Aha!" moment. The distance $d$ has completely vanished. It is no longer part of the equation.

This is the elegance of physics and mathematics—finding the invariant in a changing system.

### Phase 4: The Final Calculation

Now, we simply plug in our values. We have $H_1 = 20	ext{ m}$ and $H_2 = 80	ext{ m}$.

$$
\frac{1}{h} = \frac{1}{20} + \frac{1}{80}
$$

To add these, we find a common denominator, which is $80$. We rewrite $\frac{1}{20}$ as $\frac{4}{80}$:

$$
\frac{1}{h} = \frac{4}{80} + \frac{1}{80} = \frac{5}{80}
$$

Simplifying the fraction $\frac{5}{80}$ gives us $\frac{1}{16}$. Therefore, $\frac{1}{h} = \frac{1}{16}$, which means **$h = 16	ext{ m}$**.

### Conclusion

We have solved it. The intersection point is exactly **$16	ext{ meters}$** above the ground.

This problem teaches us that in JEE Advanced, the most powerful tool is not just calculation, but the ability to identify the underlying geometric symmetry. Whenever you see lines connecting tops to feet, remember the reciprocal sum rule.

It is a shortcut that saves time and builds confidence. Keep practicing, keep visualizing, and keep falling in love with the logic behind the numbers.

Animated Solution for Mathematics - Trigonometry: Two vertical poles of heights, 20m and 80m stand a part on a horizontal plane. The height (in meters) of the point of intersection of the lines joining the top of each pole to the foot of the other, from this horizontal plane is :

Select Answer:

Visualized Solution

Visualizing the Poles

Connecting Tops to Foot

The Intersection Point

The Standard Formula

Substituting the Values

Finding a Common Denominator

Adding the Fractions

Solving for $h$

Final Conclusion

The Sigma Insight: Heights and Distances

The Geometry of Elegance

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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