Question

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

Then $\sec 60$ is equal to:

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

Answer 1

Answer: (A)

$2$

Consider an equilateral $△ABC$, where each side $=2a$.
From $A$, draw $AD\perp BC$,

so $BD=DC=a$, and $\angle ADB={90}^o,\angle BAD={30}^o$.
From right angle $△ADB$, we have $AD=\sqrt{{AB}^2-{BD}^2}=\sqrt{{\left(2a\right)}^2-a^2}$

$=\sqrt{{4a}^2-a^2}=\sqrt{{3a}^2}=\sqrt{3}a$.

$\sec {60}^o=\frac{hypotenuse}{base}=\frac{AB}{BD}=\frac{2a}{a}=2$.

So, option A that is $2$ is correct.