Question

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

Answer 1

Hint:
Here, we will first construct an equilateral triangle with a Perpendicular bisector. Then by using the properties of an equilateral triangle, properties of the Perpendicular bisector and Pythagoras theorem we will find the lengths of all the sides of the triangle. We will use the trigonometric ratio to find the value of the given trigonometric ratio at an angle.

Formula Used:
Trigonometric Ratio: $$\tan \theta ={\displaystyle \frac{\mathrm{O}\mathrm{p}\mathrm{p}\mathrm{o}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{e}}{\mathrm{A}\mathrm{d}\mathrm{j}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}}}$$

Complete step by step solution:
We are given a Trigonometric ratio of $$\tan {30}^{\circ }$$.
Now, we will construct an equilateral triangle ABC.
We will consider an equilateral triangle ABC with the sides $$AB=BC=CA=2a$$
Now, we will draw a perpendicular $$AD$$ to $$BC$$, which divides $$BC$$ into two equal parts $$BD=DC$$.
Therefore, we get
$$BD={\displaystyle \frac{BC}{2}}$$
$$\Rightarrow BD={\displaystyle \frac{2a}{2}}$$
Dividing the terms, we get
$$\Rightarrow BD=a$$
We know that in an equilateral triangle, each angle measures $${60}^{\circ }$$.
Since $$AD$$ is a perpendicular bisector, $$\mathrm{\angle }A$$ is divided into two equal angles. i.e.
$$\mathrm{\angle }BAD={\displaystyle \frac{\mathrm{\angle }A}{2}}=\mathrm{\angle }DAC$$
$$\Rightarrow \mathrm{\angle }BAD={\displaystyle \frac{{60}^{\circ }}{2}}=\mathrm{\angle }DAC$$
Dividing the terms, we get
$$\Rightarrow \mathrm{\angle }BAD={30}^{\circ }=\mathrm{\angle }DAC$$

Since $$\mathrm{\Delta }ABD=\mathrm{\Delta }ACD$$, we get$$AD$$ is a common side to both the triangles.
$$\mathrm{\angle }ADB={90}^{\circ }=\mathrm{\angle }ADC$$
We know that $$\mathrm{\Delta }ABC\cong \mathrm{\Delta }ADC$$
Now, by using the Pythagoras theorem in $$\mathrm{\Delta }ADB$$, we will find the length of$$AD$$.
$$AD=\sqrt{A{B}^2-B{D}^2}$$
$$\Rightarrow AD=\sqrt{{\left(2a\right)}^2-{\left(a\right)}^2}=\sqrt{4a^2-a^2}$$
Subtracting the terms, we get
$$\Rightarrow AD=\sqrt{3a^2}$$
$$\Rightarrow AD=\sqrt{3}a$$
Now in $$\mathrm{\Delta }ADB$$, using the formula $$\tan \theta ={\displaystyle \frac{\mathrm{O}\mathrm{p}\mathrm{p}\mathrm{o}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{e}}{\mathrm{A}\mathrm{d}\mathrm{j}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}}}$$, we get
$$\Rightarrow \tan {30}^{\circ }={\displaystyle \frac{BD}{AD}}$$
Now, by substituting the values, we get
$$\Rightarrow \tan {30}^{\circ }={\displaystyle \frac{BD}{AD}}$$
$$\Rightarrow \tan {30}^{\circ }={\displaystyle \frac{a}{\sqrt{3}a}}$$
$$\Rightarrow \tan {30}^{\circ }={\displaystyle \frac{1}{\sqrt{3}}}$$

Therefore, the value of$$\tan {30}^{\circ }$$ is $${\displaystyle \frac{1}{\sqrt{3}}}$$.

Note:
We know that an Equilateral Triangle is a triangle having all the sides of a triangle equal. Pythagoras theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides. Trigonometric Ratios of a Particular angle are the ratios of the sides of a right angled triangle with respect to any of its acute angle. Trigonometric ratio and Pythagoras theorem is applicable only in the case of a Right Angle triangle. We should remember that an equilateral triangle should be constructed along with a Perpendicular bisector to form a Right angle triangle in an equilateral triangle.