Question

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

Then, $\tan {60}^o$ is equal to

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

Answer 1

Answer: (C)

$\sqrt{3}$

Consider an equilateral $△ABC$ with each side $=2a$.
So each angle is $60$.
From $A$, draw $\perp AD$ to $BC$,

so $BD=DC=a$ and $\angle ADB={90}^o,\angle BAD={30}^o$.
So, for $△ADB$, we have $AD=\sqrt{{AB}^2-{BD}^2}$

$=\sqrt{{\left(2a\right)}^2-a^2}=\sqrt{4a^2-a^2}=\sqrt{3a^2}=\sqrt{3}a$.
$\therefore \tan {60}^o=\frac{AD}{BD}=\frac{\sqrt{3}a}{a}=\sqrt{3}$