Question

If √3y=3x+6, then find the angle made by the line with positive x-axis.
Options are:-
i)120°
ii)60°
iii)30°
iv)150°

Answer 1

Answer:

$given \: equation \: of \: line \: is \\ \sqrt{3} y = 3x + 6 \\ y = \frac{3}{ \sqrt{3} } x + \frac{6}{ \sqrt{3} } \\ y = \sqrt{3} x + 2 \sqrt{3 } \\ comparing \: it \: with </p><p>\: slope \: intercept \: form \: of \: line \:</p><p> that \: is \: y = mx \: + c \: </p><p>we \: find \: \\ m = \sqrt{3} \: and \: c = 2 \sqrt{3} \\ where \: m \: is \: slope \: of \: line \\ since \: m \: = \tan \alpha \\ therefore \\ tan \alpha = \sqrt{3} = \tan {60}^{0} \\ \alpha = {60}^{0} \\ hence \: the \: angle \: made \: \\ by \: the \: line \: with \: positive \\ \: x-axis \: is \: {60}^{0}.$