There are three basic notable properties in a right triangle when its angle equals to $30$ degrees.

- The length of opposite side is equal to half of the length of hypotenuse.
- The length of adjacent side is equal to $\small \sqrt{3}/{2}$ times of the length of hypotenuse.
- The third angle of right triangle is $\small 60^°$.

$\Delta POQ$ is a right triangle and its angle is $30^°$. Assume, the length of hypotenuse is equal to $d$.

Then, the length of opposite side is exactly equal to half of the length of the hypotenuse.

$Length \, of \, Opposite \, side$ $\,=\,$ $\dfrac{Length \, of \, Hypotenuse}{2}$

$\implies$ $Length \, of \, Opposite \, side$ $\,=\,$ $\dfrac{d}{2}$

The lengths of opposite side and hypotenuse are known. They can be used to calculate the length of adjacent side (base) mathematically by Pythagorean Theorem.

${OP}^2 \,=\, {PQ}^2 + {OQ}^2$

$\implies d^2 \,=\, {\Bigg(\dfrac{d}{2}\Bigg)}^2 + OQ^2$

$\implies d^2 \,-\, {\Bigg(\dfrac{d}{2}\Bigg)}^2 \,=\, OQ^2$

$\implies OQ^2 \,=\, d^2 \,-\, {\Bigg(\dfrac{d}{2}\Bigg)}^2$

$\implies OQ^2 \,=\, d^2 \,-\, \dfrac{d^2}{4}$

$\implies OQ^2 \,=\, d^2 \Bigg[1 \,-\, \dfrac{1}{4}\Bigg]$

$\implies OQ^2 \,=\, d^2 \Bigg[\dfrac{3}{4}\Bigg]$

$\implies OQ^2 \,=\, \Bigg[\dfrac{3}{4}\Bigg]d^2$

$\implies OQ \,=\, \sqrt{\Bigg[\dfrac{3}{4}\Bigg]d^2}$

$\implies OQ \,=\, \dfrac{\sqrt{3}}{2}d$

$\implies OQ \,=\, \dfrac{\sqrt{3}}{2} \times d$

$\implies OQ \,=\, \dfrac{\sqrt{3}}{2} \times OP$

$\therefore \,\, Length \, of \, Adjacent \, side \,=\, \dfrac{\sqrt{3}}{2} \times Length \, of \, Hypotenuse$

The properties of right angled triangle can also be proved geometrically by constructing an example triangle with an angle of $30$ degrees.

- Firstly, draw a straight line horizontally from point $\small R$.
- Use protractor and draw a straight line from point $\small R$ but it should make an angle of $\small 30$ degrees with horizontal line.
- Take compass and a ruler, then set compass to $\small 10 \, \normalsize cm$. After that, draw an arc on $\small 30$ degrees line from point $\small R$. They both get intersected at point $\small S$.
- Draw a perpendicular line to horizontal line from point $\small S$ and it intersects the line at point $\small T$. Thus, a right triangle ($\small \Delta SRT$) is constructed geometrically.

Let us study the geometrical relations between sides of a right triangle when its angle is $30^°$.

The length of opposite side (perpendicular) equals to half of the length of hypotenuse when angle equals to $30$ degrees.

Now, take a ruler and measure the length of the opposite side ($\small \overline{ST}$). You observe that the length of opposite side $\small \overline{ST}$ is $5 \, cm$ exactly.

Actually, it is half of the length of the hypotenuse. It is possible only when the angle of a right triangle equals to $30^°$.

So, remember that the length of opposite side is always equal to half of the length of the hypotenuse when the angle of right angled triangle is $30^°$.

The length of adjacent side (base) equals to $\dfrac{\sqrt{3}}{2}$ times of the length of hypotenuse when angle equals to $30$ degrees.

Measure the length of the adjacent side $\small \overline{RT}$ by ruler and you observe that the length of the adjacent side is equal to $8.65 \, cm$. Later, calculate the length of the adjacent side in theoretical method to compare difference between them.

$Length \, of \, Adjacent \, side$ $\,=\,$ $\dfrac{\sqrt{3}}{2} \times Length \, of \, Hypotenuse$

$\implies$ $Length \, of \, Adjacent \, side \,=\, \dfrac{\sqrt{3}}{2} \times 10$

$\implies$ $Length \, of \, Adjacent \, side \,=\, 8.66025\ldots$

$\,\,\, \therefore \,\,\,\,\,\,$ $Length \, of \, Adjacent \, side \,\approx\, 8.66 \, cm$

Theoretically, the length of adjacent side is $8.66 \, cm$ approximately and geometrically, it is $8.65 \, cm$. They both are equal approximately in mathematics.

It proves that the length of adjacent side is equal to $\dfrac{\sqrt{3}}{2}$ times of length of hypotenuse when angle of right triangle is equal to $30^{°}$.

List of most recently solved mathematics problems.

Jul 04, 2018

Limit (Calculus)

Evaluate $\displaystyle \large \lim_{x \,\to\, \tan^{-1}{3}} \normalsize {\dfrac{\tan^2{x}-2\tan{x}-3}{\tan^2{x}-4\tan{x}+3}}$

Jun 23, 2018

Limit (Calculus)

Evaluate $\displaystyle \large \lim_{x \to 0} \normalsize \dfrac{e^{x^2}-\cos{x}}{x^2}$

Jun 22, 2018

Integral Calculus

Evaluate $\displaystyle \int \dfrac{1+\cos{4x}}{\cot{x}-\tan{x}} dx$

Jun 21, 2018

Limit

Evaluate $\displaystyle \large \lim_{x \to \infty} \normalsize {\sqrt{x^2+x+1}-\sqrt{x^2+1}}$

Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers.
Know more

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.