The acute angle between the lines $\left(x^2+y^2\right) \sin \theta+2 x y=0$ is
The acute angle between the lines $\left(x^2+y^2\right) \sin \theta+2 x y=0$ is
$\theta$
$\frac{\pi}{2}+\theta$
$\frac{\pi}{2}-\theta$
$\frac{\theta}{2}$
Solution
We have $\left(x^2+y^2\right) \sin \theta+2 x y=0$
$\therefore \mathrm{a}=\sin \theta, \mathrm{b}=\sin \theta$ and $\mathrm{h}=1$ and let $\alpha$ be the acute angle between the given lines.
Then $\tan \alpha=\frac{\left|2 \sqrt{(1)^2-(\sin \theta)(\sin \theta)}\right|}{\sin \theta+\sin \theta}$
$\begin{aligned}
& =\frac{\left|2 \sqrt{1-\sin ^2 \theta}\right|}{2 \sin \theta}=\frac{2 \cos \theta}{2 \sin \theta}=\cot \theta=\tan \left(\frac{\pi}{2}-\theta\right) \\
& \therefore \alpha=\frac{\pi}{2}-\theta
\end{aligned}$