Two poles, AB of length

Question:

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

  1. $x^{2}+2(a+2 b) x-b(a+b)=0$

  2. $x^{2}+2(a+2 b) x+a(a+b)=0$

  3. $x^{2}-2 a x+b(a+b)=0$

  4. $x^{2}-2 a x+a(a+b)=0$


Correct Option: , 3

Solution:

$\tan \theta=\frac{1}{2}$

$\tan (\theta+\alpha)=\frac{\mathrm{X}}{\mathrm{b}}, \tan \alpha=\frac{\mathrm{X}}{\mathrm{a}+\mathrm{b}}$

$\Rightarrow \frac{1}{2}+\frac{x}{a+b}$

$\Rightarrow \frac{\frac{1}{2}+\frac{x}{a+b}}{1-\frac{1}{2} \times \frac{x}{a+b}}=\frac{x}{b}$

$\Rightarrow x^{2}-2 a x+a b+b^{2}=0$

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