# Bibhorr Formula

## 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
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:

 $\ds \text{Bibhorr leek }$ $=$ $\ds \dfrac {(45).(AC) } {90}$ $\text {(3)}: \quad$ $\ds \leadsto \ \$ $\ds \text{ Bibhorr leek }$ $=$ $\ds \dfrac {AC} 2$

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

 $\ds \text{Bibhorr leek }$ $=$ $\ds \dfrac {AC + AB - AB} 2$ $\text {(4)}: \quad$ $\ds \leadsto \ \$ $\ds \text{ Bibhorr leek }$ $=$ $\ds \dfrac {AC} 2$

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.

$\blacksquare$