Question

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

Then, $\cos {60}^o=$

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

Answer 1

Answer: (B)

$\frac{1}{2}$

Consider an equilateral $△ABC$ with each side $=2a$.

Each angle of $△ABC$ is ${60}^o$.

From $A$, draw $AD\perp BC$, then $BD=DC=a$.

Also, $\angle ADB={90}^o,\angle BAD={30}^o$.

From right angled $△ADB$, we have

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

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

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

and hypotenuse$=AB=2a$

$\cos {60}^o=\frac{BD}{AB}=\frac{a}{2a}=\frac{1}{2}$

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