# Bibhorr Formula

## Contents

## Formula

Consider a right triangle $\triangle$ $ABC$ such that $AB$, $BC$ and $AC$ are the shortest, medium and longest sides respectively. Here, $AC$ is the hypotenuse.

Now, for this triangle the angle opposite $BC$ (*Bibhorr angle*) is given as:

- $\text{Bibhorr angle } = \dfrac {90 \left({AC + BC - AB}\right)^2} {BC^2 + 1.5 AC \left({AC + BC - AB}\right)}$ (in degrees)

or:

- $\text{Bibhorr angle } = \dfrac {\dfrac{π}{2} \left({AC + BC - AB}\right)^2} {BC^2 + 1.5 AC \left({AC + BC - AB}\right)}$ (in radians)

The above equation is the **Bibhorr formula**

## Syllabary

In Bibhorrmetry, the sides of right triangle are notated through Hindi syllabary as follows:

- hypotenuse $AC$ is notated as
*श्र* - Medium side $BC$ is represented by
*लं* - Shortest side is written as
*छ* - Bibhorr angle as
*बि*

The original equation employing above syllabary is given as:

## Proof

- Assumption

The product of Bibhorr angle and hypotenuse equals the product of $90 \degrees$ and Bibhorr leek.

Mathematically, the law is represented as:

- $(1): \quad \text{Bibhorr angle } \times \text{ hypotenuse } 90 \times \text{ Bibhorr leek}$

This assumption is referred to as Bibhorr Law.

Here, Bibhorr leek is the side of an imaginary symmetrical right triangle

- Consider the following propositions

When $AB = BC$ then:

- $(P.1): \quad \text{ Bibhorr leek} = \dfrac {AC + BC - AB} 2$

But when $AB \ne BC$ then:

- $(P.2): \quad \text{ Bibhorr leek} < \dfrac {AC + BC - AB} 2$

From $(1)$, the equation for Bibhorr leek can be deduced as:

$(2): \quad \text{ Bibhorr leek} = \dfrac {(BibhorrAngle).(hypotenuse) } {90}$

When $AB = BC$ then the angle ratio for that triangle is $2:1:1$.

This means that Bibhorr angle equals $45 \degrees$.

So $(2)$ becomes:

\(\displaystyle \text{Bibhorr leek }\) | \(=\) | \(\displaystyle \dfrac {(45).(AC) } {90}\) | $\quad$ | $\quad$ | |||||||||

\((3):\quad\) | \(\displaystyle \leadsto \ \ \) | \(\displaystyle \text{ Bibhorr leek }\) | \(=\) | \(\displaystyle \dfrac {AC} 2\) | $\quad$ | $\quad$ |

In such case proposition $(P.1)$ becomes:

\(\displaystyle \text{Bibhorr leek }\) | \(=\) | \(\displaystyle \dfrac {AC + AB - AB} 2\) | $\quad$ | $\quad$ | |||||||||

\((4):\quad\) | \(\displaystyle \leadsto \ \ \) | \(\displaystyle \text{ Bibhorr leek }\) | \(=\) | \(\displaystyle \dfrac {AC} 2\) | $\quad$ | $\quad$ |

From $(4)$ and $(3)$, proposition $(P.1)$ is proved.

Recalling proposition $(P.2)$

When $AB \ne BC$ then $\text{ Bibhorr leek} < \dfrac {AC + BC - AB} 2$

In order to balance the inequality a special variable $ज $ is added to the denominator:

- $(5): \quad \text{ Bibhorr leek} = \dfrac {AC + BC - AB} {2 + ज }$

- Special variable

The variable equals as follows:

- $ज = \paren {BC^2 - \dfrac {BC^2 AB^2} {AC^2} }^\dfrac1 2 - \paren {\dfrac {AC + BC - AB} 2}$

By further computing the equation is reduced to:

- $(6): \quad ज = {\dfrac {BC^2} {AC (AC + BC - AB)} } - \dfrac 1 2$

Putting $(6)$ in $(5)$, we have

- $(7): \quad \text{ Bibhorr leek } = \dfrac {AC \left({AC + BC - AB}\right)^2} {BC^2 + 1.5 AC \left({AC + BC - AB}\right)}$

This result is Bibhorr Leek Theorem.

Putting $(7)$ in $(1)$, we get:

- $\text{ Bibhorr angle } = \dfrac {90 \left({AC + BC - AB}\right)^2} {BC^2 + 1.5 AC \left({AC + BC - AB}\right)}$ (in degrees)

This is the desired result.

Thus, deriving Bibhorr formula.

$\blacksquare$