Special Right Triangle Calculator
Solve 30°-60°-90° and 45°-45°-90° special right triangles with exact radical formatting ($1 : 1 : \sqrt{2}$ and $1 : \sqrt{3} : 2$), step-by-step geometric proofs, and interactive SVG diagrams.
Special Right Triangle Rules & Side Ratios
Special right triangles possess fixed geometric angle configurations that allow exact side length calculations without trigonometric tables. In a 45°-45°-90° triangle, sides follow the 1 : 1 : √2 ratio. In a 30°-60°-90° triangle, sides follow the 1 : √3 : 2 ratio.
Exact Ratio Rules (45°-45°-90° & 30°-60°-90°)
Special right triangles are geometric shortcuts that eliminate the need for calculator approximations by expressing all side lengths as exact radical multiples of a single known parameter $x$:
45°-45°-90° Triangle Ratio
- Legs: $a = b = x$
- Hypotenuse: $c = x\sqrt{2}$
- Area: $\text{Area} = \frac{x^2}{2}$
30°-60°-90° Triangle Ratio
- Short Leg (opp 30°): $a = x$
- Long Leg (opp 60°): $b = x\sqrt{3}$
- Hypotenuse (opp 90°): $c = 2x$
45°-45°-90° Isosceles Right Triangle Derivation
The 45°-45°-90° triangle is formed by drawing a diagonal across any square of side length $x$. By the Pythagorean theorem $a^2 + b^2 = c^2$:
Because the base angles are congruent ($45^\circ = 45^\circ$), it is the only right triangle that is also isosceles.
30°-60°-90° Equilateral Bisection Derivation
The 30°-60°-90° triangle is derived by cutting an equilateral triangle with side length $2x$ in half with an altitude:
This proves why the hypotenuse is exactly double the shorter leg, and the longer leg is $\sqrt{3}$ times the shorter leg.
Connection to the Trigonometric Unit Circle Coordinates
| Angle (θ) | Radians | $\sin(\theta)$ | $\cos(\theta)$ | $\tan(\theta)$ |
|---|---|---|---|---|
| 30° | $\pi / 6$ | $1/2$ | $\sqrt{3}/2$ | $\sqrt{3}/3$ |
| 45° | $\pi / 4$ | $\sqrt{2}/2$ | $\sqrt{2}/2$ | $1$ |
| 60° | $\pi / 3$ | $\sqrt{3}/2$ | $1/2$ | $\sqrt{3}$ |
Step-by-Step Worked Numerical Solutions
Find the legs of a 45°-45°-90° triangle with hypotenuse $c = 10$.
1. Formula: $c = a\sqrt{2} \implies a = \frac{c}{\sqrt{2}}$.
2. Substitute $c = 10$: $a = \frac{10}{\sqrt{2}}$.
3. Rationalize denominator: $a = \frac{10 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$.
Both legs $a = b = 5\sqrt{2} \approx 7.071$.
Common Pitfalls & Radical Rationalization
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