The acute angle included between the lines $x \sin \theta-y \cos \theta=5$ and $x \sin \propto-y \cos…
The acute angle included between the lines $x \sin \theta-y \cos \theta=5$ and
$x \sin \propto-y \cos \propto+11=0$ is
$|\theta-\propto|$
$\frac{\pi}{4}$
$\frac{\pi}{3}$
$\theta+\propto$
Solution
Slope of line $x \sin \theta-y \cos \theta=5$ is $m_{1}=\frac{\sin \theta}{\cos \theta}=\tan \theta$ Slope of line $x \sin \alpha-y \sin \alpha+11=0$ is $m_{2}=\frac{\sin \alpha}{\cos \alpha}=\tan \alpha$ Let $\beta$ be the angle between the lines
$\begin{aligned}
\tan \beta &=\left|\frac{\tan \theta-\tan \alpha}{1+\tan \alpha \tan \alpha}\right| \Rightarrow \tan \beta=\tan (\theta-\alpha) \\
\beta &=|\theta-\alpha|
\end{aligned}$