Question

In $△MNK$, $\angle N={90}^o$, $\angle M={60}^o$ and $MN=6$. Then find $NK$.

• A. $6$ units
• B. $6\sqrt{3}$ units
• C. $2\sqrt{3}$ units
• D. $5\sqrt{3}$ units

Answer 1

The correct option is B $6\sqrt{3}$ units

In $△MNK,\angle N={90}^o,\angle M={60}^o$ and $MN=6$

In triangle $MNK,$

$\angle M+\angle N+\angle K={180}^o$

$\Rightarrow {90}^o+{60}^o+\angle K={180}^o$

$\Rightarrow \angle K={30}^o$ ..... Since$\angle MNK={90}^o$

$\Rightarrow \tan {30}^o=\frac{MN}{NK}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{6}{NK}$

$\Rightarrow NK=6\sqrt{3}$ units

Hence, option B is correct.