A vertical pole fixed to the horizontal ground is divided in the ratio 3 : 7 by a mark on it with lo

Q: Wait, why can't we just use similar triangles to solve this?

That is a common instinct! However, similar triangles require the angles to be the same in both triangles. Here, the triangles share a base, but their heights are different ($3\ell$ vs $10\ell$), and the angles subtended are $\alpha$ and $2\alpha$. Since $\alpha \neq 2\alpha$, the triangles are not similar. We must use trigonometry to relate the sides and angles.

Q: Why did we define the total height as $10\ell$ instead of just $H$?

Great question! The problem gives us a ratio of $3:7$. By setting the parts as $3\ell$ and $7\ell$, we ensure the ratio is satisfied while keeping the algebra clean. If we used $H$, we would have to work with fractions like $\frac{3}{10}H$ and $\frac{7}{10}H$ from the start, which makes the equations messier. Using $\ell$ is a strategic choice to simplify the calculation.

Q: Is it possible for $\ell$ to be zero? Can we just cancel it out?

Mathematically, if $\ell = 0$, the equation holds, but physically, a pole of height zero doesn't exist! Since we are solving for a physical dimension, we know $\ell > 0$. Therefore, it is perfectly safe to divide both sides by $\ell$ to simplify our equation.

Question

A vertical pole fixed to the horizontal ground is divided in the ratio $3 : 7$ by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground $18	ext{ m}$ away from the base of the pole, then the height of the pole (in meters) is :

Options:

(A) $12\sqrt{15}$

(B) $12\sqrt{10}$

(C) $8\sqrt{10}$

(D) $6\sqrt{10}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a flat, horizontal stretch of ground. In front of you stands a vertical pole divided into two distinct segments by a mark.

The lower part is shorter, and the upper part is longer, in a ratio of $3:7$. You stand exactly $18	ext{ m}$ away from the base, looking up.

The lower part of the pole and the upper part of the pole subtend the exact same angle at your eye. This is a classic JEE Advanced problem that tests your ability to translate a physical scenario into a precise mathematical model.

### The Blueprint

First, we must visualize the geometry. Let the total height of the pole be $10\ell$. The mark divides this into a lower segment of $3\ell$ and an upper segment of $7\ell$.

We have a point on the ground, let us call it $P$, which is $18	ext{ m}$ from the base of the pole. If we draw lines of sight from $P$ to the mark and from $P$ to the top of the pole, we form two right-angled triangles.

Let the angle subtended by the lower part be $\alpha$. The problem states that the upper part also subtends an angle $\alpha$. Therefore, the entire pole subtends an angle of $\alpha + \alpha = 2\alpha$ at point $P$.

### The Trigonometric Bridge

Now, we apply our trigonometric toolkit. For the smaller triangle, which corresponds to the lower part of the pole, we have:

$$	an \alpha = \frac{3\ell}{18} = \frac{\ell}{6}$$

Next, we look at the larger triangle, which encompasses the entire pole. The total height is $10\ell$, and the base remains $18	ext{ m}$. The angle subtended by the whole pole is $2\alpha$.

$$	an 2\alpha = \frac{10\ell}{18} = \frac{5\ell}{9}$$

We use the double angle identity to bridge these expressions:

$$	an 2\alpha = \frac{2	an \alpha}{1 - 	an^2 \alpha}$$

### The Algebraic Resolution

Now, we substitute our expressions into the identity:

$$\frac{5\ell}{9} = \frac{2(\ell/6)}{1 - (\ell/6)^2}$$

The right-hand side simplifies as follows:

$$\frac{5\ell}{9} = \frac{\ell/3}{1 - \ell^2/36} = \frac{\ell/3}{(36 - \ell^2)/36} = \frac{12\ell}{36 - \ell^2}$$

Since $\ell$ represents a physical length, it cannot be zero. Dividing both sides by $\ell$, we obtain:

$$\frac{5}{9} = \frac{12}{36 - \ell^2}$$

Cross-multiplying gives $5(36 - \ell^2) = 108$, which simplifies to $180 - 5\ell^2 = 108$. Rearranging, we find $5\ell^2 = 72$, or $\ell^2 = 14.4$.

### Final Calculation

The question asks for the total height of the pole, which we defined as $H = 10\ell$. Substituting our value for $\ell$:

$$H = 10\sqrt{14.4} = \sqrt{100 	imes 14.4} = \sqrt{1440}$$

Simplifying the radical:

$$H = \sqrt{144 	imes 10} = 12\sqrt{10}	ext{ m}$$

The **total height of the pole** is **$12\sqrt{10}	ext{ m}$**.

Select Answer:

Visualized Solution

Visualizing the Pole and the Mark

Defining the Subtended Angles

Applying Trigonometry: $tan α$

Applying Trigonometry: $tan 2 α$

The Double Angle Formula

Substitution and Setup

Simplifying the Expression

Solving for $ℓ^{2}$

Finding the Total Height

Final Simplification

Summary and Key Takeaway

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Blueprint

The Trigonometric Bridge

The Algebraic Resolution

Final Calculation

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