Question

In right angle $△ABD,\angle B={60}^o,\angle A={30}^o$, then $\cot {60}^o=$

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

Answer 1

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

in $△ABD$

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

Therefore, Answer is $\frac{1}{\sqrt{3}}$