Question

The given figure represents the lines $y=x+1$ and $y=\sqrt{3}-1$. Write down the angels which the lines make with the positive direction of the x-axis. Hence, determine $\theta$.

Answer 1

Let line $x+1=y$ intersect $x$ axis at $A$

and other line at $B$, also another line intersect $x$ axis at $C$,

Now for $y=x+1$, slope$=1$, so, $\angle BAC={\tan }^{-1}\left(1\right)={45}^{\circ }$

and for $y=\sqrt{3}x-1$ slope$=\sqrt{3}$, so, $\angle BAC={\tan }^{-1}\left(\sqrt{3}\right)={60}^{\circ }$

Using the external angle property in triangle $ABC$ we get,

$\angle BAC=\angle ABC=\angle BCX⟹{45}^{\circ }+\theta ={60}^{\circ }⟹\theta ={15}^{\circ }$