Question

In a right angled $△ABD,\angle B={60}^o$ and $\angle A={30}^o$.

Then, $\cot {60}^o=$

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

Answer 1

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

Consider an equilateral $△ABC$, with each side $=2a$.
So each angle of $△ABD$ is ${60}^o$.

From $A$, draw $AD\perp BC$, so $BD=DC=a$.

Also, $\angle BAD={30}^o$ and $\angle ADB={90}^o$.

From right angled $△ADB,AD=\sqrt{{AB}^2-{BD}^2}$

$=\sqrt{{\left(2a\right)}^2-a^2}=\sqrt{{4a}^2-a^2}=\sqrt{{3a}^2}=\sqrt{3}a$.

$\cot {60}^o=\frac{base}{perpendicular}=\frac{BD}{AD}=\frac{a}{\sqrt{3}a}=\frac{1}{\sqrt{3}}$.

So $\frac{1}{\sqrt{3}}$ is correct.