Right Triangle Solver

Solve right-angled triangles instantly using the Pythagorean theorem, trigonometry ($\sin, \cos, \tan$), and Euclidean geometric identities. Complete with interactive SVG geometry diagram.

Euclidean Geometry & Trigonometric Engine
units
Length of first leg side
units
Length of second leg side
Precision for lengths & trigonometric values
Geometric Proportion Diagram
Diagram drawn to scale with vertex angles & hypotenuse altitude $h_c$
Leg a (Vertical)
3.0000
Leg b (Base)
4.0000
Hypotenuse c
5.0000
Angle α (Opp a)
36.8699°
0.6435 rad
Angle β (Opp b)
53.1301°
0.9273 rad
Angle γ (Right)
90.0000°
π / 2 rad
Area (K)
6.0000
Perimeter (P)
12.0000
Altitude (h_c)
2.4000
Inradius (r)
1.0000
r = (a + b - c) / 2
Circumradius (R)
2.5000
R = c / 2
Semi-Perimeter (s)
6.0000
s = (a + b + c) / 2
Leg Ratio (a / b)
0.7500
tan(α) aspect

Trigonometric Ratios for Acute Angles (α & β)

Exact trigonometric definitions derived from side length ratios.

Trig Function Angle α Value Formula (α) Angle β Value Formula (β)

Euclidean Trigonometry, Right Triangle Geometry & Analytic Metric Solvers

The right-angled triangle (a planar polygon with three vertices, three sides, and one internal angle measuring exactly $\gamma = 90^\circ$ or $\frac{\pi}{2}\text{ radians}$) occupies a central position in Euclidean geometry, structural engineering, surveying, satellite triangulation, and vector physics. Because one angle is fixed by definition at $90^\circ$, determining all remaining geometric properties requires only two independent parameters (at least one of which must be a side length).

The Fundamental Geometric Theorems

Designating the two orthogonal legs as $a$ and $b$, the hypotenuse opposite the right angle as $c$, and the acute angles opposite sides $a$ and $b$ as $\alpha$ and $\beta$ respectively:

1. The Pythagorean Theorem

$$a^2 + b^2 = c^2 \implies c = \sqrt{a^2 + b^2}, \quad a = \sqrt{c^2 - b^2}, \quad b = \sqrt{c^2 - a^2}$$

2. Complementary Angle Postulate

Because the interior angle sum of any Euclidean planar triangle is $180^\circ$:

$$\alpha + \beta + 90^\circ = 180^\circ \implies \alpha + \beta = 90^\circ = \frac{\pi}{2}\text{ rad}$$

3. Primary Trigonometric Ratios

$$\sin(\alpha) = \frac{a}{c} = \cos(\beta), \quad \cos(\alpha) = \frac{b}{c} = \sin(\beta), \quad \tan(\alpha) = \frac{a}{b} = \cot(\beta)$$

Analytical Solution Matrix for Input Parameter Pairs

Depending on the two known inputs provided, the system executes one of four analytic branches:

  1. Two Legs Known ($a, b$): $$c = \sqrt{a^2 + b^2}, \quad \alpha = \arctan\left(\frac{a}{b}\right), \quad \beta = 90^\circ - \alpha$$
  2. Leg & Hypotenuse Known ($a, c$ where $c > a$): $$b = \sqrt{c^2 - a^2}, \quad \alpha = \arcsin\left(\frac{a}{c}\right), \quad \beta = 90^\circ - \alpha$$
  3. Leg & Adjacent/Opposite Angle Known ($a, \alpha$): $$\beta = 90^\circ - \alpha, \quad c = \frac{a}{\sin(\alpha)}, \quad b = \frac{a}{\tan(\alpha)}$$
  4. Hypotenuse & Acute Angle Known ($c, \alpha$): $$\beta = 90^\circ - \alpha, \quad a = c \cdot \sin(\alpha), \quad b = c \cdot \cos(\alpha)$$

Secondary Geometric Invariants

  • Planar Area ($K$): $$K = \frac{1}{2} \cdot a \cdot b = \frac{1}{2} \cdot c \cdot h_c$$
  • Perimeter ($P$): $$P = a + b + c$$
  • Altitude to Hypotenuse ($h_c$): $$h_c = \frac{a \cdot b}{c}$$
  • Inradius ($r$): Radius of the inscribed circle: $$r = \frac{a + b - c}{2} = \frac{a \cdot b}{a + b + c}$$
  • Circumradius ($R$): By Thales's Theorem, the hypotenuse forms the circle's diameter: $$R = \frac{c}{2}$$

Comprehensive Right Triangle Example

Consider a right triangle with legs $a = 6.0$ and $b = 8.0$:

  1. Hypotenuse ($c$): $$c = \sqrt{6.0^2 + 8.0^2} = \sqrt{36 + 64} = \sqrt{100} = \mathbf{10.0}$$
  2. Angles ($\alpha$ and $\beta$): $$\alpha = \arctan\left(\frac{6}{8}\right) = \arctan(0.75) \approx \mathbf{36.87^\circ} \; (0.6435\text{ rad})$$ $$\beta = 90^\circ - 36.87^\circ = \mathbf{53.13^\circ} \; (0.9273\text{ rad})$$
  3. Area & Perimeter: $$K = \frac{6 \cdot 8}{2} = \mathbf{24.0}, \qquad P = 6 + 8 + 10 = \mathbf{24.0}$$
  4. Altitude & Inradius: $$h_c = \frac{6 \cdot 8}{10} = \mathbf{4.8}, \qquad r = \frac{6 + 8 - 10}{2} = \mathbf{2.0}$$
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