Bibhorr Formula
This theorem is false: Trigonometric functions are transcendental. Although it can be used for approximation in some conditions, it is not accurate.
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Formula
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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:
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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:
This is not proved.
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- $ज = \paren {BC^2 - \dfrac {BC^2 AB^2} {AC^2} }^\dfrac1 2 - \paren {\dfrac {AC + BC - AB} 2}$
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.
$\blacksquare$
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