Find the angle that the line $y = \sqrt{3}\,x + 1$ makes with the positive $x$-axis.

Question

Accepted Answer

$60^\circ$. The gradient of a line equals the tangent of the angle it makes with the positive $x$-axis. Here $	an	heta = \sqrt{3}$, so $	heta = 60^\circ$. The angle $30^\circ$ corresponds to a gradient of $\dfrac{1}{\sqrt{3}}$.

Answer

A) $45^\circ$

Answer

B) $60^\circ$

Answer

C) $30^\circ$

Answer

D) $90^\circ$

Find the angle that the line $y = \sqrt{3}\,x + 1$ makes with the positive $x$-axis.

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