JEE Mains · Maths · STD 11 - Trigonometrical equations
Two poles, \(\mathrm{AB}\) of length \(a\) metres and \(\mathrm{CD}\) of length \(\mathrm{a}+\mathrm{b}(\mathrm{b} \neq \mathrm{a})\) metres are erected at the same horizontal level with bases at \(\mathrm{B}\) and \(\mathrm{D} .\) If \(\mathrm{BD}=\mathrm{x}\) and \(\tan \angle\,ACB=\frac{1}{2}\), then:
- A \(x^{2}+2(a+2 b) x-b(a+b)=0\)
- B \(x^{2}+2(a+2 b) x+a(a+b)=0\)
- C \(x^{2}-2 a x+b(a+b)=0\)
- D \(x^{2}-2 a x+a(a+b)=0\)
Answer & Solution
Correct Answer
(C) \(x^{2}-2 a x+b(a+b)=0\)
Step-by-step Solution
Detailed explanation
\(\tan \theta=\frac{1}{2}\) \(\tan (\theta+\alpha)=\frac{x}{b}, \tan \alpha=\frac{x}{a+b}\) \(\Rightarrow \frac{1}{2}+\frac{x}{a+b}\) \(\Rightarrow \frac{\frac{1}{2}+\frac{x}{a+b}}{1-\frac{1}{2} \times \frac{x}{a+b}}=\frac{x}{b}\) \(\Rightarrow x^{2}-2 a x+a b+b^{2}=0\)
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