Question

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

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$2$

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

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

Now $\csc 30=\frac{AB}{BD}=\frac{2a}{a}=2$.