Unknowns in Triangle

A triangle can be defined by its 3 side lengths or its 3 vertex angles, its area, its perimeter or a mixture of these values. These values are very dependent on each other, so if some are unknown, the unknowns of the triangle can therefore be calculated from the known values.

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

Calculation formula for the 3 angles (unknown values), 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 $$ \beta = \gamma $$

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. Take into account these equations:

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