Two vertical poles are 150 m apart and the height of one is three times that of the other. If from t

Q: Why can't we just use the sine rule? It seems easier.

You absolutely could! The sine rule is a powerful tool. However, in problems involving right-angled triangles and angles of elevation, the tangent ratio is often the most direct path because it links the height (the unknown) directly to the base (the known). Using the sine rule would introduce the hypotenuse, which is an extra variable you'd have to solve for or eliminate. Always aim for the path of least resistance!

Q: Wait, why do we reject the negative root here?

Great catch! In pure algebra, $h^2 = k$ gives $h = \pm\sqrt{k}$. But remember, we are dealing with physical reality here. $h$ represents the height of a pole. A physical length cannot be negative. Therefore, we discard the negative root because it has no physical meaning in this context.

Q: Is it always true that the product of the tangents of complementary angles is 1?

Yes, absolutely! If $\alpha + \beta = 90^\circ$, then $\beta = 90^\circ - \alpha$. Therefore, $\tan \alpha \cdot \tan \beta = \tan \alpha \cdot \tan(90^\circ - \alpha) = \tan \alpha \cdot \cot \alpha = 1$. This is a beautiful property that often appears in JEE problems to simplify calculations. Keep this trick in your toolkit!

Question

Two vertical poles are 150 m apart and the height of one is three times that of the other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is:

Options:

(A) 25

(B) $20\sqrt{3}$

(C) 30

(D) $25\sqrt{3}$

Sigma Percentile · Accepted Answer

### The Geometry of Elevation

Imagine you are standing on a perfectly flat, expansive plain. In the distance, two vertical poles rise toward the sky. They are separated by a distance of $150$ m.

One is a towering giant, three times the height of its companion. You stand exactly at the midpoint between them, looking up at their summits. This is the classic setup of a JEE trigonometry problem—a scenario that tests not just your ability to calculate, but your ability to visualize the hidden relationships between physical objects.

### The Setup: Visualizing the Poles

Let us define our variables with precision. We denote the height of the shorter pole as $h$. Consequently, the taller pole, being three times as tall, has a height of $3h$.

The total distance between the poles is $150$ m. Since you are standing at the midpoint, you are exactly 
$$d = \frac{150}{2} = 75 	ext{ m}$$ 
away from the base of each pole. This symmetry is our first gift; it simplifies the geometry significantly.

### The Trigonometric Bridge

Now, consider the angles of elevation. You look up at the shorter pole at an angle $	heta$. You then turn to the taller pole, and the problem tells us the angle of elevation is complementary to the first. This means the second angle is $90^{\circ} - 	heta$.

In the right-angled triangle formed by the shorter pole, we use the tangent ratio: 
$$	an 	heta = \frac{	ext{Opposite}}{	ext{Adjacent}} = \frac{h}{75}$$

For the taller pole, we do the same: 
$$	an(90^{\circ} - 	heta) = \frac{3h}{75}$$

### The Elegant Elimination

Here is where the magic happens. We have two equations, but we have an unknown angle $	heta$ that we don't actually need to find. We only care about $h$.

We recall the fundamental co-function identity: $	an(90^{\circ} - 	heta) = \cot 	heta$. Substituting this into our second equation, we get:
$$\cot 	heta = \frac{3h}{75}$$

Now, we have two beautiful expressions: $	an 	heta = \frac{h}{75}$ and $\cot 	heta = \frac{3h}{75}$. If we multiply these two equations together, the $	heta$ terms will cancel out because $	an 	heta \cdot \cot 	heta = 1$. This is the 'Aha!' moment that separates the novice from the master.

### The Final Reveal

Multiplying the two equations, we get:
$$1 = \left( \frac{h}{75} ight) \cdot \left( \frac{3h}{75} ight)$$

$$1 = \frac{3h^{2}}{75^{2}}$$

Cross-multiplying gives us $3h^{2} = 75^{2}$. Dividing by $3$, we find $h^{2} = \frac{75^{2}}{3}$. Taking the square root of both sides, we arrive at:
$$h = \frac{75}{\sqrt{3}}$$

To match our answer with the standard form, we rationalize the denominator by multiplying the numerator and denominator by $\sqrt{3}$:
$$h = \frac{75\sqrt{3}}{3} = 25\sqrt{3} 	ext{ m}$$

And there it is. The height of the shorter pole is **$25\sqrt{3}$ meters**. It is a clean, elegant result derived from the simple, yet powerful, interplay of trigonometric identities. Remember, in JEE, the most complex problems often yield to the most fundamental principles if you look at them with the right perspective.

Select Answer:

Visualized Solution

Visualizing the Physical Setup

Locating the Observer

Defining the Angles of Elevation

Trigonometry in the First Triangle

Trigonometry in the Second Triangle

Applying Co-function Identity

The Strategy to Eliminate $θ$

Multiplying the Equations

Simplifying the Expression

Isolating $h^{2}$

Taking the Square Root

Rationalizing the Denominator

The Sigma Insight: Heights and Distances

The Geometry of Elevation

The Setup

The Trigonometric Bridge

The Elegant Elimination

The Final Reveal

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