Question

Find the value of $\tan {60}^{\circ }$ geometrically.

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

Answer 1

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

Consider an equilateral $\Delta PQR$. Since each angle in an equilateral triangle.

$\therefore \angle P=\angle Q=\angle R={60}^0$

Draw a perpendicular $PS$ from $P$ to $QR$.

Now, In $\Delta PQS$ and $\Delta PRS$

$\angle PSQ=\angle PSR\left[eachof{90}^0\right]$

$\angle PQS=\angle PRS\left[eachof{60}^0\right]$

$PS=PS\left[common\right]$

$\therefore \Delta PQS\cong \Delta PRS$$(A.S.A.congruancy)$

Let sides $PQ=QR=RP=2x$

Then, $QS=SR=\frac{QR}{2}=\frac{2x}{2}=x$

Now, In $\Delta PSQ$

$P{Q}^2=Q{S}^2+S{P}^2$ By Pythagoras theorem

${\left(2x\right)}^2=x^2+P{S}^2$

$P{S}^2=3x^2$

$PS=\sqrt{3}x$

Now, In $\Delta PSQ$

$\tan Q=\frac{PS}{QS}$

$\tan {60}^0=\frac{\sqrt{3}x}{x}$

$\tan {60}^0=\sqrt{3}$

Hence, this is the answer.