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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

## Triangle Unknown Values Calculator

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.

### 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 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:

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

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

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

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

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

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

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:

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

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

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

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

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

### 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 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\$$

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

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

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

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:

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

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

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

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

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

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

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:

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

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

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

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

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

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

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:

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

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

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

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

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

### 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 angle $$\alpha$$ are equals $$b = c$$

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

If $$b = 3$$ and $$\beta = \frac{\pi}{6}$$, Then $$c = 3$$ and $$\gamma = \frac{\pi}{6}$$

### How to simplify calculations knowing the triangle is rectangle?

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 $$\mathcal{A} = \frac{ab}{2}$$

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