If $m=1$ is the slope of a line $L$,then the product of the slopes of non-parallel lines which are inclined at an angle of $60^{\circ}$ with $L$ is

The acute angle between two straight lines passing through the point $M(-6, -8)$ and the points in which the line segment $2x + y + 10 = 0$ enclosed between the coordinate axes is divided in the ratio $1 : 2 : 2$ in the direction from the point of its intersection with the $x$-axis to the point of intersection with the $y$-axis is:

Statement $(A)$ : The line $2x + y + 6 = 0$ is perpendicular to the line $x - 2y + 5 = 0$ and the second line passes through $(1, 3)$.
Reason $(R)$ : The product of the slopes of perpendicular lines is $-1$.

The acute angle between the lines given by $y - \sqrt{3}x + 1 = 0$ and $\sqrt{3}y - x + 7 = 0$ is (in $^{\circ}$)

The obtuse angle between the lines $y = -2$ and $y = x + 2$ is .....$^o$

Let $\theta_1$ be the angle between two lines $2x + 3y + c_1 = 0$ and $-x + 5y + c_2 = 0$,and $\theta_2$ be the angle between two lines $2x + 3y + c_1 = 0$ and $-x + 5y + c_3 = 0$,where $c_1, c_2, c_3$ are any real numbers.
Statement-$1$: If $c_2$ and $c_3$ are proportional,then $\theta_1 = \theta_2$.
Statement-$2$: $\theta_1 = \theta_2$ for all $c_2$ and $c_3$.