Question

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

Then $\csc {60}^o$ is equal to:

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

Answer 1

Answer: (B)

$\frac{2}{\sqrt{3}}$

Consider an equilateral $△ABC$ with each side $=2a$.
Each angle is ${60}^o$.
From $A$, draw $AD\perp BC$, then $BD=DC=a$.
Also, $\angle ADB={90}^o,\angle BAD={30}^o$.
From right angle $△ADB$, we have
$AD=\sqrt{{AB}^2-{BD}^2}=\sqrt{{\left(2a\right)}^2-a^2}$

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

$\csc 60=\frac{1}{\sin 60}=\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}$
So $\frac{2}{\sqrt{3}}$ is correct.