Question

Find the value of $\tan {{60}^{\circ }}$ geometrically.
A). $\sqrt{5}$
B). $\sqrt{3}$
C). $-\sqrt{3}$
D). $\sqrt{2}$

Answer 1

Hint: In this question, we have to find the tangent of ${{60}^{\circ }}$. Therefore, we should try to construct an appropriate figure from which we can find out the tangent of ${{60}^{\circ }}$from its definition.

Complete step-by-step solution
Let the figure below represent an equilateral triangle CBD with each side equal to 1cm and let CA represent the perpendicular from C to BD. As the triangle is equilateral, $\angle CBD=\angle CDB=\angle BCD={{60}^{\circ }}$……………..(1.1)

We should note that by RHS criterion, two right angled triangles are said to be congruent ($\cong $) if the length of the hypotenuse and one side is equal in both the triangles…………………………….. (1.2)
In $\Delta CBA$ and $\Delta CDA$, we find that
$\angle CAD=\angle CAB={{90}^{\circ }}$
$\angle CBA=\angle CDA={{60}^{\circ }}$
$CB=CD\text{ (as }\Delta \text{CBD is equilateral)}$
Therefore, by the RHS criterion stated in equation (1.2), $\Delta CBA\cong \Delta CDA........................(1.3)$
Now, the corresponding sides in congruent triangles are equal. Therefore, from equation (1.3), as the triangles are congruent
$BA=AD.....................(1.4)$
Therefore, as $\Delta CBD$ is equilateral of side length 1cm and from equation (1.4), we get
$\begin{align}
  & BD=BA+AD=2BA=1cm \\
 & \Rightarrow BA=AD=\dfrac{1}{2}cm........................(1.5) \\
\end{align}$
Also, as $\Delta CAD$ is an equilateral triangle, by Pythagoras theorem,
\[\begin{align}
  & \text{hypotenuse}{{\text{e}}^{2}}\text{=Sum of square of sides} \\
 & \Rightarrow \text{C}{{\text{D}}^{2}}=\text{C}{{\text{A}}^{2}}+A{{\text{D}}^{2}} \\
 & \Rightarrow CA=\sqrt{C{{D}^{2}}-A{{D}^{2}}}=\sqrt{1-{{\left( \dfrac{1}{2} \right)}^{2}}}=\sqrt{\dfrac{3}{4}}..................(1.6) \\
\end{align}\]
Now, tangent of an angle in a right-angled triangle is given by
$\tan \left( \theta \right)=\dfrac{\text{length of the side facing angle }\theta }{\text{length of the side adjacent to angle }\theta }.....................(1.7)$
Therefore, using equation (1.5) in $\Delta CAD$, as $\angle CDA={{60}^{\circ }}$ from equation (1.1), (1.6) and (1.7), we get
$\tan (\theta )=\dfrac{CA}{AD}=\dfrac{\sqrt{\dfrac{3}{4}}}{\dfrac{1}{2}}=2\times \dfrac{\sqrt{3}}{2}=\sqrt{3}$
Thus, the value of $\tan ({{60}^{\circ }})$ is found geometrically to be $\sqrt{3}$. Thus, option (b) is the correct answer.

Note: In the above problem, we have shown the value of $\tan {{60}^{\circ }}$ as $\sqrt{3}$ geometrically but as you can see that it is a multiple-choice question so if you remember the value of $\tan {{60}^{\circ }}$ then you don’t have to solve it geometrically. The point to be noted that you should know the sine, cosine, tangent of the following angles: ${{0}^{\circ }},{{30}^{\circ }},{{45}^{\circ }},{{60}^{\circ }},{{90}^{\circ }}$. It will help you in solving various trigonometric questions.
In this problem also, if you would have known that $\tan {{60}^{\circ }}$ is equal to $\sqrt{3}$ then in the final answer when you are solving it geometrically if you are not getting $\sqrt{3}$ then you know that you have made a mistake.