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Two poles, A B of length a metres and C D 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}=x and \tan \angle \mathrm{ACB}=\frac{1}{2}, then :
Option: 1 x^{2}-2 \mathrm{a} x+\mathrm{a}(\mathrm{a}+\mathrm{b})=0
Option: 2 x^{2}+2(\mathrm{a}+2 \mathrm{~b}) x-\mathrm{b}(\mathrm{a}+\mathrm{b})=0
Option: 3 x^{2}+2(\mathrm{a}+2 \mathrm{~b}) x+\mathrm{a}(\mathrm{a}+\mathrm{b})=0
Option: 4 x^{2}-2 \mathrm{a} x+\mathrm{b}(\mathrm{a}+\mathrm{b})=0

Answers (1)

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\angle ACB= \angle ACD-\angle BCD\\

\Rightarrow \tan\angle ACB=\tan\left ( \angle ACD-\angle BCD \right )\\

\tan\angle ACD= \frac{x}{b}, \tan\angle BCD= \frac{x}{a+b}\\

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

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

\Rightarrow x^{2}-2ax+b\left ( a+b \right )= 0

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Kuldeep Maurya

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