If the angle $2 \theta$ is acute, then the acute angle between the pair of straight lines $$ \begin{aligned}…

If the angle $2 \theta$ is acute, then the acute angle between the pair of straight lines $$ \begin{aligned} & x^2(\cos \theta-\sin \theta)+2 x y \cos \theta \\ & \quad+y^2(\cos \theta+\sin \theta)=0, \text { is } \end{aligned} $$
  1. $2 \theta$
  2. $\frac{\theta}{2}$
  3. $\frac{\theta}{3}$
  4. $\theta$

Solution

We have $ x^2(\cos \theta-\sin \theta)+2 x y \cos \theta $ We know that, $ \begin{aligned} \tan \alpha & =\frac{2 \sqrt{h^2-a b}}{a+b} \\ \tan \alpha & =\frac{2 \sqrt{\cos ^2 \theta-(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)}}{(\cos \theta+\sin \theta)+(\cos \theta-\sin \theta)} \\ & =\frac{2 \sqrt{\cos ^2 \theta-\cos ^2 \theta+\sin ^2 \theta}}{2 \cos \theta}=\tan \theta \\ \Rightarrow \alpha & =\theta \end{aligned} $

Asked in: AP EAMCET 2002

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