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 $$BD = x$$ and $$\tan \angle ACB = \frac{1}{2}$$, then:

Step 1: Establish a Coordinate System

Let us place the base of the first pole, $$B$$, at the origin $$(0,0)$$.

Since the bases $$B$$ and $$D$$ are on the same horizontal level and separated by a distance $$x$$ ($$BD = x$$), the coordinates of $$D$$ are $$(x, 0)$$.

Now, we can find the coordinates of the tops of the poles:

• Pole $$AB$$ has length $$a$$, so point $$A$$ is at $$(0, a)$$.
• Pole $$CD$$ has length $$a+b$$, so point $$C$$ is at $$(x, a+b)$$.

Step 2: Determine the Slopes of Lines CA and CB

The angle $$\angle ACB$$ is the angle formed between the line segments $$CA$$ and $$CB$$. Let us find the slopes ($$m$$) of both these lines using the formula $$m = \frac{y_2 - y_1}{x_2 - x_1}$$.

Slope of line $$CA$$ ($$m_1$$):

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

Slope of line $$CB$$ ($$m_2$$):

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

Step 3: Apply the Angle Between Two Lines Formula

The tangent of the angle $$\theta$$ between two lines with slopes $$m_1$$ and $$m_2$$ is given by:

$$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$$

Substitute $$m_1$$, $$m_2$$, and the given $$\tan \angle ACB = \frac{1}{2}$$ into the formula:

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

Step 4: Simplify and Solve the Equation

Simplify the numerator and the denominator on the right side:

• Numerator: $$\frac{a+b - b}{x} = \frac{a}{x}$$
• Denominator: $$1 + \frac{b(a+b)}{x^2} = \frac{x^2 + b(a+b)}{x^2}$$

Put them back together:

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

Now, cross-multiply to form the final quadratic equation:

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

Hence, the answer is option C.