Question

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

Then $\tan {30}^o$ is equal to

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

Answer 1

Answer: (B)

$\frac{1}{\sqrt{3}}$

Consider an equilateral $△ABC$, with each side $=2a$.

So, each angle is ${60}^o$. From $A$, draw $AD\perp BC$.

So, $BD=DC=a,\angle ADB={90}^o,\angle BAD={30}^o$.

So, for right angled $△ADB$, we have

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

$\tan 30=\frac{BD}{AD}=\frac{a}{\sqrt{3}a}=\frac{1}{\sqrt{3}}$.

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