Question

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

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

Answer 1

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

Consider an equilateral triangle with each side $=2a$.

So each angle of $△ABC$ is ${60}^o$.

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

From right angled triangle $△ADB$, we have

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

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

Now, in right angled $△ADB$, base $BD=a,AD=\sqrt{3}a$

and hypotenuse $=AB=2a$
$\sin {60}^o=\frac{AD}{AB}=\frac{\sqrt{3}\times a}{2\times a}=\frac{\sqrt{3}}{2}$

So, $\frac{\sqrt{3}}{2}$ is correct.