# Perimeter of Right Triangle by Tangents

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BobDH
Perimeter of Right Triangle by Tangents

Proposition: The perimeter of a right triangle equals the altitude to the hypotenuse multiplied by the sum of the Tan's of 1/2 the acute∠’s, divided by the product of the Tan's of 1/2 the acute ∠’s

You may say this is a non-problem, and you would be right. Simply add up the lengths of all 3 sides.

However, the proposition seeks to reveal just one of many mysterious and profound ways of mathematics.

EXAMPLE:
To demonstrate, let the 8 15 17 Pythagorean triangle be our selection.

1. Calculating the altitude to hypotenuse we arrive at 8x15/17 = 7.058823529. See following link:

https://en.wikipedia.org/wiki/Geometric_mean_theorem?wprov=sfla1

2. Using the triangle in the link below, calculate the acute ∠’s A & B.

3. Arctan∠A = 8/15 = 28.07248694°.

4.1/2(28.07248694°)=14.03624347°

5.Tan14.03624347°= 0.250 . . .

6. Arctan∠B=15/8 = 61.92751306°.

7.1/2(61.92751306°)=30.96375653°

NOTE: 14.03624347°+ 30.96375653°= 45°. Since the sum of acute ∠'s in the 8 15 17 triangle equals 90°, then 1/2 their sum equals 45°. See Items 4 & 7.

8. Tan 30.96375653°= 0.600 . . .

9. Sum of Tan's of 1/2 (∠'s A-B) =
0.250 + 0.600 = 0.850 . . .

10. Alt to Hyp is: 7.058823529. See Item 1.

11. 0.850 x 7.058823529=6.000. See Item 9.

12. Prod of Tan's of 1/2 (∠'s A-B) = 0.250 x 0.600 = 0.150. See Items 5 & 8.

13. 6.000 / 0.150 = 40.000 Perim of Triangle. See Items 11 & 12.

14. 8 + 15 + 17 = 40.000 Perim of Triangle.

AMAZING!!!

fitzmerl duron

aba nice! :-D

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