Tool to find unknowns in a triangle. Resolving triangle equations allows to solve all unknowns in the triangle knowing only 2 or 3 caracteristic values.

Unknowns in Triangle - dCode

Tag(s) : Geometry

dCode is free and its tools are a valuable help in games, puzzles and problems to solve every day!

You have a problem, an idea for a project, a specific need and dCode can not (yet) help you? You need custom development? *Contact-me*!

Sponsored ads

Tool to find unknowns in a triangle. Resolving triangle equations allows to solve all unknowns in the triangle knowing only 2 or 3 caracteristic values.

Considering the three sides \( a \), \( b \) and \( c \) are known in the triangle.

Calculation formula for the 3 angles, the area and the perimeter are:

$$ \alpha = \arccos\left( \frac{b^2+c^2-a^2}{2bc} \right) $$

$$ \beta = \arccos\left( \frac{c^2+a^2-b^2}{2ca} \right) $$

$$ \gamma = \arccos\left( \frac{a^2+b^2-c^2}{2ab} \right) $$

$$ \mathcal{A} = \frac14\sqrt{(a+b+c)(a+b-c)(-a+b+c)(a-b+c)} $$

$$ \mathcal{P} = a+b+c $$

Considering one angle \( \gamma \) and its adjacent sides \( a \) and \( b \) are known in the triangle.

Calculation formula for the 2 other angles, the opposite side, the area and the perimeter are:

$$ c = \sqrt{a^2+b^2-2ab\cos\gamma} $$

$$ \alpha = \frac\pi2 - \frac\gamma2 + \arctan\left(\frac{a-b}{(a+b)\tan\frac\gamma2}\right) $$

$$ \beta = \frac\pi2 - \frac\gamma2 - \arctan\left(\frac{a-b}{(a+b)\tan\frac\gamma2}\right) $$

$$ \mathcal{A} = \frac12 ab\sin\gamma $$

$$ \mathcal{P} = a+b+\sqrt{a^2+b^2-2ab\cos\gamma} $$

Considering 1 angle \( \beta \), its adjacent sides \( c \) and the opposite side \( b \) are known in the triangle.

If \( \beta \) is acute and \( b < c \) then calculation formula for the 2 other angles, the last adjacent side, the area and the perimeter are:

$$ a = c\cos\beta-\sqrt{b^2-c^2\sin^2\beta} $$

$$ \gamma = \pi-\arcsin\left(\frac{c\sin\beta}b\right) $$

$$ \alpha = -\beta + \arcsin\left(\frac{c\sin\beta}b\right) $$

$$ \mathcal{A} = \frac 12 c\left(\sqrt{b^2-c^2\sin^2\beta}-c\cos\beta\right)\sin\beta $$

$$ \mathcal{P} = c\cos\beta-\sqrt{b^2-c^2\sin^2\beta}+b+c $$

If \( \beta \) is not acute or if \( b >= c \) then calculation formula for the 2 other angles, the last adjacent side, the area and the perimeter are:

$$ a = \sqrt{b^2-c^2\sin^2\beta}+c\cos\beta $$

$$ \alpha = \pi-\beta-\arcsin\left(\frac{c\sin\beta}b\right) $$

$$ \gamma = \arcsin \left(\frac{c\sin\beta}b\right) $$

$$ \mathcal{A} = \frac 12c\left(\sqrt{b^2-c^2\sin^2\beta}+c\cos\beta\right)\sin\beta $$

$$ \mathcal{P} = \sqrt{b^2-c^2\sin^2\beta}+c\cos\beta+b+c $$

Considering the 2 angles \( \alpha \) and \( \beta \) and their common side \( c \) are known in the triangle.

Calculation formula for the 2 other sides, the last angle, the area and the perimeter are:

$$ a = \frac {c\sin\alpha}{\sin(\alpha+\beta)} $$

$$ b = \frac {c\sin\beta}{ \sin(\alpha+\beta)} $$

$$ \gamma = \pi-\alpha-\beta\ $$

$$ \mathcal{A} = \frac12 c^2 \, \frac{\sin\alpha\sin\beta}{\sin(\alpha+\beta)} $$

$$ \mathcal{P} = \frac {c ( \sin\alpha + \sin\beta )}{ \sin(\alpha+\beta)} + c $$

Considering the 2 angles \( \alpha \) and \( \beta \) and one of their non common side \( a \) are known in the triangle.

Calculation formula for the 2 other sides, the last angle, the area and the perimeter are:

$$ b = \frac{a\sin\beta}{\sin\alpha} $$

$$ c = \frac{a\sin(\alpha+\beta)}{\sin\alpha} $$

$$ \gamma = \pi-\alpha-\beta $$

$$ \mathcal{A} = \frac12 a^2 \, \frac{\sin(\alpha+\beta)\sin\beta}{\sin\alpha} $$

$$ \mathcal{P} = a + \frac{a(\sin\beta+\sin(\alpha+\beta))}{\sin\alpha} $$

Considering the area \( \mathcal{A} \), the angle \( \gamma \) and one adjacent side \( a \) are known in the triangle.

Calculation formula for the 2 other sides, the other 2 angles and the perimeter are:

$$ b = \frac{2\mathcal{A}}{a\sin\gamma} $$

$$ c = \frac{1}{a} \sqrt{a^2-\frac{4 \mathcal{A}}{\tan{\gamma}}+\frac{4 \mathcal{A}^2}{a^2\sin{\gamma}^2}} $$

$$ \alpha = \frac{1}{2} \left(\pi -\gamma +2 \arctan{\frac{a-\frac{2 \mathcal{A}}{a \sin\gamma}}{\left(a+\frac{2 \mathcal{A}}{a\sin\gamma}\right)\tan{\frac{\gamma}{2}}}}\right) $$

$$ \beta = \frac{1}{2} \left(\pi -\gamma -2 \arctan{\frac{a-\frac{2 \mathcal{A}}{a \sin\gamma}}{\left(a+\frac{2 \mathcal{A}}{a\sin\gamma}\right)\tan{\frac{\gamma}{2}}}}\right) $$

$$ \mathcal{P} = \frac{1}{a} \left( a^2 + \frac{2\mathcal{A}}{\sin\gamma} + \sqrt{a^2-\frac{4 \mathcal{A}}{\tan{\gamma}}+\frac{4 \mathcal{A}^2}{a^2\sin\gamma^2}} \right) $$

Considering the area \( \mathcal{A} \), the angle \( \alpha \) and one adjacent side \( a \) are known in the triangle.

Calculation formula for the 2 other sides, the other 2 angles and the perimeter are:

$$ b = \frac{1}{\sqrt{2}}\sqrt{a^2+\frac{4\mathcal{A}}{\tan\alpha}+a\sqrt{a^2-\frac{16\mathcal{A}^2}{a^2}+\frac{8\mathcal{A}}{\tan\alpha}}} $$

$$ c = \frac{1}{\sqrt{2}}\sqrt{a^2+\frac{4\mathcal{A}}{\tan\alpha}-a\sqrt{a^2-\frac{16\mathcal{A}^2}{a^2}+\frac{8\mathcal{A}}{\tan\alpha}}} $$

$$ \beta = \arcsin\left(\frac{2\sqrt{2}\mathcal{A}}{a\sqrt{a^2+\frac{4\mathcal{A}}{\tan\alpha}-a\sqrt{a^2-\frac{16\mathcal{A}^2}{a^2}+\frac{8\mathcal{A}}{\tan\alpha}}}}\right) $$

$$ \gamma = \arcsin\left(\frac{2\sqrt{2}\mathcal{A}}{a\sqrt{a^2+\frac{4\mathcal{A}}{\tan\alpha}+a\sqrt{a^2-\frac{16\mathcal{A}^2}{a^2}+\frac{8\mathcal{A}}{\tan\alpha}}}}\right) $$

$$ \mathcal{P} = a+\frac{1}{\sqrt{2}}\left( \sqrt{a^2+\frac{4\mathcal{A}}{\tan\alpha}+a\sqrt{a^2-\frac{16\mathcal{A}^2}{a^2}+\frac{8\mathcal{A}}{\tan\alpha}}} +\sqrt{a^2+\frac{4\mathcal{A}}{\tan\alpha}-a\sqrt{a^2-\frac{16\mathcal{A}^2}{a^2}+\frac{8\mathcal{A}}{\tan\alpha}}} \right) $$

Considering the area \( \mathcal{A} \) and the two sides \( b \) and \( c \) are known in the triangle.

Calculation formula for the last side, the 3 angles and the perimeter are:

$$ a = \sqrt{b^2+c^2+2 \sqrt{b^2 c^2-4 \mathcal{A}^2}} $$

$$ \alpha = \arccos\left(-\frac{\sqrt{b^2 c^2-4 \mathcal{A}^2}}{b c}\right) $$

$$ \beta = \arccos\left(\frac{2 c^2+2 \sqrt{2+b^2 c^2-4 \mathcal{A}}}{2 c \sqrt{b^2+c^2+2 \sqrt{b^2 c^2-4 \mathcal{A}^2}}}\right) $$

$$ \gamma = \arccos\left(\frac{2 b^2+2 \sqrt{b^2 c^2-4 \mathcal{A}}}{2 b \sqrt{b^2+c^2+2 \sqrt{b^2 c^2-4 \mathcal{A}^2}}}\right) $$

$$ \mathcal{P} = \sqrt{b^2+c^2+2 \sqrt{b^2 c^2-4 \mathcal{A}^2}} + b + c $$

Considering the triangle is isosceles in \( A \).

The 2 sides forming the angle \( \alpha \) are equals $$ b = c $$

The 2 angles that are adjacent to the third side \( a \) are equals $$ \beta = \gamma $$

Example: If \( b = 3 \) and \( \beta = \frac{\pi}{6} \), Then \( c = 3 \) and \( \gamma = \frac{\pi}{6} \)

Considering the triangle is rectangle in \( C \).

The angle \( \gamma \) is right $$ \gamma = 90° = \frac\pi2 $$

The sum of the 2 other angles is equal to 90° $$ \alpha + \beta = 90° = \frac\pi2 $$

The Pythagorean theorem can be applied $$ a^2 + b^2 = c^2 $$

The area of the triangle can be simplified as $$ \mathcal{A} = \frac{ab}{2} $$

Considering the triangle is equilateral.

The 3 sides are equal $$ a = b = c $$

The 3 angles are equal to 60° $$ \alpha = \beta = \gamma = 60° = \frac\pi3 $$

The perimeter can be simplified as $$ \mathcal{P} = 3a = 3b = 3c $$

dCode retains ownership of the source code of the script Unknowns in Triangle online. Except explicit open source licence (indicated Creative Commons / free), any algorithm, applet, snippet, software (converter, solver, encryption / decryption, encoding / decoding, ciphering / deciphering, translator), or any function (convert, solve, decrypt, encrypt, decipher, cipher, decode, code, translate) written in any informatic langauge (PHP, Java, C#, Python, Javascript, Matlab, etc.) which dCode owns rights will not be given for free. To download the online Unknowns in Triangle script for offline use on PC, iPhone or Android, ask for price quote on contact page !

- How to solve knowing the 3 sides?
- How to solve knowing 1 angle and the 2 adjacent sides?
- How to solve knowing 1 angle, the opposite side and 1 adjacent side?
- How to solve knowing 2 angles and the common side?
- How to solve knowing 2 angles and 1 non-common side?
- How to solve knowing the area, 1 angle and 1 adjacent side?
- How to solve knowing the area, 1 angle and the opposite side?
- How to solve knowing the area and 2 sides?
- How to simplify calculations knowing the triangle is isosceles?
- How to simplify calculations knowing the triangle is rectangle?
- How to simplify calculations knowing the triangle is equilateral?

triangle,area,perimeter,point,angle,side,isosceles,rectangle,equilateral,vertex,unknown,equation

Source : https://www.dcode.fr/unknowns-triangle

© 2018 dCode — The ultimate 'toolkit' to solve every games / riddles / geocaches. dCode

Feedback