Jadhav Isosceles Formula

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Jadhav's Isosceles Formula gives a equation to evaluate area of any isosceles triangle with know one angle and length of equal sides with trigonometric ratios. Derived by Jyotiraditya Jadhav

Statement

In any isosceles triangle let the length of equal sides be "s" and the angle formed between both the sides be $\theta$. then the area of the complete triangle can be found by Jadhav Isosceles Formula as below:

$[{sin (\theta/2)}{cos( \theta /2)}{s2}]$

Nomenclature

  1. $\theta$/ Theta : The angle formed between the two equal sides of the triangle
  2. s/s2 : The length of the equal sides of the triangle (square of the length)
  3. Sin : The ratio used in triangle to compare the opposite side (to the angle $\theta$) of the triangle with respect to the hypotenuse of the triangle.
  4. Cos : The ratio used in triangle to compare the adjacent side (to the angle $\theta$) of the triangle with respect to the hypotenuse of the triangle

A Brief about Trigonometry and the necessary functions needed to understand the formula

Trigonometry (from Greek trigōnon, "triangle" and metron, "measure") is a branch of mathematics that studies relationships between side lengths and angles of triangles. The field emerged in the Hellenistic world during the 3rd century BC from applications of geometry to astronomical studies. The Greeks focused on the calculation of chords, while mathematicians in India created the earliest-known tables of values for trigonometric ratios (also called trigonometric functions) such as sine.

About Sin and Cosine Functions:

If the angle θ is given, then all sides of the right-angled triangle are well-defined up to a scaling factor. This means that the ratio of any two side lengths depends only on θ. these two ratios define two functions of θ, which are the trigonometric functions comparing opposite side to θ and adjacent side to θ with the hypotenuse of the right-angled triangle.

Sin = Opposite Side / Hypotenuse

Cosine/ Cos = Adjacent Side/ Hypotenuse

Formulas for area of an Isosceles triangle till now

  • Area = ½ × base × Height
  • A = ½[√(a2 − b2 ⁄4) × b]
  • A = ½ × b × c × sin()
  • A = [c2×sin()×sin()/ 2×sin(2π−α−β)]
  • A = ½ × a2

Derivation

  1. In Figure 1. ABC is an Isosceles triangle with AM as median, and now the area of this triangle will be $2 \times 1/2 \times MB \times AM$
  2. So the area is $MB \times AM$, now with respect to angle A ($\theta/2$) , $\sin (\theta/2)= MB/AB$ so $MB = \sin (\theta/2) \times AB$
  3. $\cos(\theta/2)= AM/AB$ , so $AM = \cos(\theta/2) \times AB$
  4. So the area becomes $MB X AM = [{\sin(\theta/2)}{\cos(\theta/2)(s2)}]$
Figure 1 : Isosceles Triangle
Isosceles Triangle





Note

  • While using the formula, we should put the value of θ as the angle formed between the two equal sides of the triangle and not any other angle, if in case only other angle is available then the angle between both the equal sides can be found by uses of Isosceles triangle Theorem : If two sides of a triangle are congruent , then the angles opposite to these sides are congruent. and also by using angle property of any triangle : All triangles angles sum up to 180
  • The Sin and Cosine ratios should be of the angle half of θ

Applications

Can be used to determine the area of an Isosceles triangle

Can be used to find the angle formed between two equal sides of the triangle given area