Question

Let $O=(0,0)$; let $A$ and $B$ be points respectively on $x$-axis and $y$-axis such that $\angle OBA={60}^{\circ }$. Let $D$ be a point in the first quadrant such that $OAD$ is an equilateral triangle. Then the slope of $DB$ is

• A. $\sqrt{3}$
• B. $\sqrt{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is D $\frac{1}{\sqrt{3}}$


In $△OBA$
$\tan {60}^{\circ }=\frac{a}{b}=\sqrt{3}\Rightarrow a=b\sqrt{3}$
Cordinates of $D=(a\cos {60}^{\circ },a\sin {60}^{\circ })=(\frac{a}{2},\frac{a\sqrt{3}}{2})=(\frac{b\sqrt{3}}{2},\frac{3b}{2})$
$m_{DB}=\frac{\frac{3b}{2}-b}{\frac{b\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}$