Question

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

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

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

Answer 1

Answer: (B)

$\frac{1}{2}$

Consider an equilateral $△ABC$, with each side $=2a$.
So, each angle of $△ABC={60}^o$.

From $A$ draw $AD\perp BC$. So $BD=DC=a$, $\angle ADB={90}^o,\angle BAD={30}^o$.

From right angled $△ADB$, we have $AD=\sqrt{{AB}^2-{BD}^2}=\sqrt{{\left(2a\right)}^2-a^2}$

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

$\sin {30}^o=\frac{BD}{AB}=\frac{a}{2a}=\frac{1}{2}$.

So, $\frac{1}{2}$ is correct.