Question

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

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

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{2}{\sqrt{3}}$
• C. $\frac{3}{\sqrt{3}}$
• D. $\frac{4}{\sqrt{3}}$

Answer 1

The correct option is B $\frac{2}{\sqrt{3}}$

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

So, each angle is ${60}^o$.

From $A$, draw $AD\perp BC$.

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

So, for right angled $△ADB$, we have

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

$\sec {30}^o=\frac{1}{\cos {30}^o}=\frac{1}{\frac{\sqrt{3}}{2}}=\frac{AB}{AD}=\frac{2}{\sqrt{3}}$.
So $\frac{2}{\sqrt{3}}$ is correct.