Two poles, AB of length a metres and CD of length a + b (b ≠ a) metres are erected at the same horiz

Q: Wait, why can't we just use the Law of Cosines in $\triangle ACB$?

You absolutely can! The Law of Cosines is a powerful tool. However, in this specific problem, calculating the lengths of $AC$ and $BC$ using the distance formula and then applying the Law of Cosines leads to a much more complex algebraic expression involving square roots. Coordinate geometry allows us to work with slopes directly, which simplifies the algebra significantly and reduces the chance of calculation errors.

Q: Why did we use the absolute value in the tangent formula?

The formula $\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$ gives the acute angle between two lines. Since the problem provides $\tan \angle ACB = \frac{1}{2}$, which is positive, we know the angle is acute. The absolute value ensures we are dealing with the correct geometric angle, though in this specific setup, the expression inside the absolute value will naturally be positive.

Q: Is there a way to solve this without coordinate geometry?

Yes, you could use trigonometry by dropping a perpendicular from $A$ to $CD$ or by using the tangent subtraction formula $\tan(\alpha - \beta)$. However, coordinate geometry is generally the most robust method for these types of problems because it provides a systematic, step-by-step framework that is less prone to geometric intuition traps.

Question

Two poles, AB of length $a$ metres and CD of length $a + b (b 
eq a)$ metres are erected at the same horizontal level with bases at B and D. If $BD = x$ and $	an \angle ACB = \frac{1}{2}$, then:

Options:

(A) $x^2 + 2(a + 2b)x - b(a + b) = 0$

(B) $x^2 + 2(a + 2b)x + a(a + b) = 0$

(C) $x^2 - 2ax + b(a + b) = 0$

(D) $x^2 - 2ax + a(a + b) = 0$

Sigma Percentile · Accepted Answer

### The Geometry of Poles: A Coordinate Odyssey

Imagine you are standing on a perfectly flat, infinite plain. In front of you, two vertical poles are erected. One is shorter, with height $a$, and the other is taller, with height $a + b$.

They are separated by a horizontal distance $x$. This is the physical reality of our problem, which serves as a classic JEE Advanced challenge testing your ability to translate physical space into the language of mathematics.

### Mapping the Terrain

To solve this, we must first build a map using a coordinate system. Let's place the base of the shorter pole, $B$, at the origin $(0,0)$.

Since the pole $AB$ is vertical and has length $a$, its top $A$ is at $(0, a)$. The base of the second pole, $D$, is at a distance $x$ from $B$, so $D$ is at $(x, 0)$.

The second pole $CD$ has length $a + b$, so its top $C$ is at $(x, a + b)$. We have now successfully translated the physical poles into four points: $A(0, a)$, $B(0, 0)$, $C(x, a + b)$, and $D(x, 0)$.

### The Slopes of Connection

Now, we focus on the angle $\angle ACB$. This angle is formed by the intersection of lines $CA$ and $CB$.

To find the tangent of this angle, we need the slopes of these two lines. Let $m_1$ be the slope of line $CA$. Using the slope formula $m = \frac{y_2 - y_1}{x_2 - x_1}$, we calculate:

$$m_1 = \frac{(a + b) - a}{x - 0} = \frac{b}{x}$$

Next, let $m_2$ be the slope of line $CB$. Using the coordinates of $C(x, a + b)$ and $B(0, 0)$, we find:

$$m_2 = \frac{(a + b) - 0}{x - 0} = \frac{a + b}{x}$$

### The Tangent Bridge

We know that the tangent of the angle $	heta$ between two lines with slopes $m_1$ and $m_2$ is given by the formula:

$$	an 	heta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} ight|$$

We are given that $	an \angle ACB = \frac{1}{2}$. Substituting our slopes into this formula, we obtain:

$$\frac{1}{2} = \frac{\frac{a + b}{x} - \frac{b}{x}}{1 + \left(\frac{b}{x}ight)\left(\frac{a + b}{x}ight)}$$

### The Algebraic Dance

Now, let's simplify this expression. The numerator simplifies to:

$$\frac{a + b - b}{x} = \frac{a}{x}$$

The denominator simplifies to:

$$1 + \frac{b(a + b)}{x^2} = \frac{x^2 + b(a + b)}{x^2}$$

Putting it all together, we have:

$$\frac{1}{2} = \frac{\frac{a}{x}}{\frac{x^2 + b(a + b)}{x^2}}$$

When we simplify this complex fraction, we get:

$$\frac{1}{2} = \frac{ax}{x^2 + b(a + b)}$$

Cross-multiplying gives us $x^2 + b(a + b) = 2ax$. Rearranging the terms, we arrive at the final quadratic equation:

**$$x^2 - 2ax + b(a + b) = 0$$**

This elegant result perfectly matches the required geometric condition. You have successfully navigated the geometry and emerged victorious!

Animated Solution for Mathematics - Straight Lines: Two poles, AB of length $a$ metres and CD of length $a + b (b \neq = a)$ metres are erected at the same horizontal level with bases at B and D. If $B D = x$ and $tan ∠ A C B = \frac{1}{2}$ , then:

Select Answer:

Visualized Solution

Visualizing the Setup

Setting up the Coordinate System

Connecting the Points

Identifying the Angle

Slope of Line $C A$ ( $m_{1}$ )

Slope of Line $C B$ ( $m_{2}$ )

The Angle Formula

Substituting the Values

Simplifying the Numerator

Simplifying the Denominator

Combining and Canceling

Cross-Multiplication

The Final Quadratic Equation

The Sigma Insight: Angle Between Two Lines

The Geometry of Poles

Mapping the Terrain

The Slopes of Connection

The Tangent Bridge

The Algebraic Dance

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