Question

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 :

• A. \(x^2 + 2(a + 2b)x - b(a+b) = 0\)
• B. \(x^2 - 2ax + a(a + b) = 0\)
• C. \(x^2 - 2ax + b(a+b) = 0\)
• D. \(x^2 + 2(a + 2b)x + a(a+b) = 0\)

Hint:

For complex angles in height and distance problems, a very useful strategy is to construct horizontal and vertical lines to form right-angled triangles. Then, express the required angle as a sum or difference of angles within these right triangles and apply the appropriate tangent sum/difference formula.

Answer 1

The Correct Option is C

Step 1: Set up a diagram and define angles.
Draw a diagram with two vertical poles AB and CD on a horizontal line BD. AB=a, CD=a+b, BD=x. Draw a line from C parallel to BD, meeting the line AB (extended if necessary) at a point P.
- We have a right-angled trapezoid ABDC (if viewed from the side).
- In our construction, P is on the vertical line through B, and CP is horizontal.
- CP = BD = x.
- The height from P to B is the same as the height of pole CD. So, BP = CD = a+b.
- The length AP is the difference in height between P and A. AP = BP - AB = (a+b) - a = b.
Step 2: Use trigonometry in the constructed right-angled triangles.
We have two right-angled triangles with vertex C: \(\triangle CPB\) and \(\triangle CPA\).
- In \(\triangle CPB\), \(\angle CPB = 90^\circ\). We can define \(\angle PCB\).
\[ \tan(\angle PCB) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BP}{CP} = \frac{a+b}{x} \] - In \(\triangle CPA\), \(\angle CPA = 90^\circ\). We can define \(\angle PCA\).
\[ \tan(\angle PCA) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AP}{CP} = \frac{b}{x} \] Step 3: Use the angle subtraction formula for tangents.
From the diagram, it is clear that \(\angle ACB = \angle PCB - \angle PCA\).
We can use the formula \(\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\).
\[ \tan(\angle ACB) = \tan(\angle PCB - \angle PCA) = \frac{\tan(\angle PCB) - \tan(\angle PCA)}{1 + \tan(\angle PCB)\tan(\angle PCA)} \] Substitute the expressions from Step 2 and the given value \(\tan(\angle ACB) = 1/2\).
\[ \frac{1}{2} = \frac{\frac{a+b}{x} - \frac{b}{x}}{1 + \left(\frac{a+b}{x}\right)\left(\frac{b}{x}\right)} \] Step 4: Simplify the expression and solve for the equation.
\[ \frac{1}{2} = \frac{\frac{a+b-b}{x}}{1 + \frac{b(a+b)}{x^2}} = \frac{\frac{a}{x}}{\frac{x^2 + b(a+b)}{x^2}} \] \[ \frac{1}{2} = \frac{a}{x} \cdot \frac{x^2}{x^2 + ab + b^2} = \frac{ax}{x^2 + ab + b^2} \] Now, cross-multiply:
\[ 1 \cdot (x^2 + ab + b^2) = 2 \cdot (ax) \] \[ x^2 + ab + b^2 = 2ax \] Rearrange to match the options:
\[ x^2 - 2ax + ab + b^2 = 0 \] \[ x^2 - 2ax + b(a+b) = 0 \] This matches option (C).