Find the angle between the lines $y - \sqrt{3}x - 5 = 0$ and $\sqrt{3}y - x + 6 = 0$. (in $^{\circ}$)

The slope of a line $L$ passing through the point $(-2, -3)$ is not defined. If the angle between the lines $L$ and $ax - 2y + 3 = 0$ $(a > 0)$ is $45^{\circ}$,then the angle made by the line $x + ay - 4 = 0$ with the positive $X$-axis in the anticlockwise direction is

The equation of the straight line passing through the point $(3, 2)$ and inclined at an angle of $60^{\circ}$ with the line $\sqrt{3} x + y = 1$ is

The angle between the lines $x = 2$ and $x - 3y = 6$ is

If two lines $x+(a-1)y=1$ and $2x+a^2y=1$ $(a \in R-\{0,1\})$ are perpendicular,then the distance of their point of intersection from the origin is

The angle between the lines $y = (2 - \sqrt{3})x + 5$ and $y = (2 + \sqrt{3})x - 7$ is.......$^o$