Find the acute angle between the lines $y = 3$ and $y = \sqrt{3}x + 9$. (in $^{\circ}$)

Consider a triangle $ABC$ having the vertices $A(1,2)$,$B(\alpha, \beta)$,and $C(\gamma, \delta)$. The angles are $\angle ABC = \frac{\pi}{6}$ and $\angle BAC = \frac{2\pi}{3}$. If the points $B$ and $C$ lie on the line $y = x + 4$,then $\alpha^2 + \gamma^2$ is equal to:

Find the equation of the line passing through the point $(3, -2)$ and making an angle of $60^{\circ}$ with the line $\sqrt{3}x + y = 1$.

If the straight lines $2x + 3y - 3 = 0$ and $x + ky + 7 = 0$ are perpendicular,then the value of $k$ is:

Two equal sides of an isosceles triangle are along $-x+2y=4$ and $x+y=4$. If $m$ is the slope of its third side,then the sum of all possible distinct values of $m$ is:

If $\alpha, \beta$ $(\alpha > \beta)$ are two values of $k$ such that the equations $2x + (3 - 2k)y + (2k + 1) = 0$ and $kx + (k - 1)y - 4 = 0$ represent two perpendicular lines,then $\alpha^2 + 2\beta =$