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Unknowns in Triangle

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.

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Unknowns in Triangle -

Tag(s) : Geometry,Mathematics

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Unknowns in Triangle

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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.

Answers to Questions

How to solve knowing the 3 sides?

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:

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

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

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

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

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

How to solve knowing 1 angle and the 2 adjacent sides?

Considering one anglehref \( \gamma \) 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:

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

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

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

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

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

How to solve knowing 1 angle, the opposite side and 1 adjacent side?

Considering 1 anglehref \( \beta \), 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:

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

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

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

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

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

How to solve knowing 2 angles and the common side?

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 anglehref, the area and the perimeter are:

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

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

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

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

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

How to solve knowing 2 angles and 1 non-common side?

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

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

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

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

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

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

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

How to solve knowing the area, 1 angle and 1 adjacent side?

Considering the area \( \mathcal{A} \), the anglehref \( \gamma \) 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:

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

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

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

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

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

How to solve knowing the area, 1 angle and the opposite side?

Considering the area \( \mathcal{A} \), the anglehref \( \alpha \) 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:

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

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

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

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

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

How to solve knowing the area and 2 sides?

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:

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

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

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

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

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

How to simplify calculations knowing the triangle is isosceles?

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

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

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

How to simplify calculations knowing the triangle is rectangle?

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

The anglehref \( \gamma \) 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 $$ \mathcal{A} = \frac{ab}{2} $$

How to simplify calculations knowing the triangle is equilateral?

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 $$ \mathcal{P} = 3a = 3b = 3c $$

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