Question

Find the value of $\tan {30^ \circ }$ geometrically.

Answer 1

Hint: We will take an equilateral triangle. Draw a perpendicular from one of the vertex to the opposite side. Find all the angles and length of all the sides of the resultant right triangle. Substitute the values in the formula of $\tan \theta = \dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}$ to find the desired value.

Complete step-by-step answer:
We will begin the question by considering an equilateral triangle \[\vartriangle ABC\] with the length of each side be $a$ units.
Also, in an equilateral triangle, each angle is of ${60^ \circ }$
Then, draw a perpendicular from $A$ on \[BC\]

Since each angle in an equilateral triangle is of ${60^ \circ }$ and $\angle ADB = {90^ \circ }$, then, find the angle $\angle BAD$ using angle sum property of a triangle.
As, it is known that the angle sum property states that, the sum of all angles in a triangle is ${180^ \circ }$
Then, in $\vartriangle ADC$, $\angle ADB + \angle BAD + \angle ABD = {180^ \circ }$
Substitute the values of the given angles to find the value of $\angle BAD$.
$
  {90^ \circ } + \angle BAD + {60^ \circ } = {180^ \circ } \\
   \Rightarrow {150^ \circ } + \angle BAD = {180^ \circ } \\
   \Rightarrow \angle BAD = {30^ \circ } \\
 $
Hence, we have,

Also, the perpendicular in an equilateral triangle divides the opposite side in two equal parts.
That is, $BD = DC$
Therefore, $BD = \dfrac{a}{2}$ , where length of each side of equilateral triangle is $a$ units.
Now, we have to find the value of $\tan {30^ \circ }$
Therefore, the base in triangle $\vartriangle ADC$ is $AD$, perpendicular is \[BD\] and the hypotenuse is \[AB\]
We will calculate the value of $AD$ using Pythagoras theorem in $\vartriangle ADC$.
$
  A{B^2} = A{D^2} + B{D^2} \\
   \Rightarrow {a^2} = A{D^2} + {\left( {\dfrac{a}{2}} \right)^2} \\
   \Rightarrow {a^2} = A{D^2} + \dfrac{{{a^2}}}{4} \\
   \Rightarrow A{D^2} = {a^2} - \dfrac{{{a^2}}}{4} \\
   \Rightarrow A{D^2} = \dfrac{{4{a^2} - {a^2}}}{4} \\
   \Rightarrow A{D^2} = \dfrac{{3{a^2}}}{4} \\
   \Rightarrow AD = \dfrac{{\sqrt 3 a}}{2} \\
 $
As it is known, $\tan \theta = \dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}$
Substitute the value of perpendicular \[BD = \dfrac{a}{2}\] and base \[AD = \dfrac{{\sqrt 3 a}}{2}\] for finding the value of $\tan {30^ \circ }$.
$
  \tan {30^ \circ } = \dfrac{{\dfrac{a}{2}}}{{\dfrac{{\sqrt 3 a}}{2}}} \\
   \Rightarrow \tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }} \\
$

Note: Many students interpret base and perpendicular incorrectly. Base is the length of the triangle which makes the required angle with the hypotenuse of the triangle. Similarly, values of $\cos \theta = \dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}}$ and $\sin \theta = \dfrac{{{\text{Opposite}}}}{{{\text{Hypotenuse}}}}$ can be calculated.