02 // Visual Math & Finance Solversc = √(a² + b²), tan(α) = b / a, Area = ½·a·b
Visual Right-Triangle Solver (SOHCAHTOA, Altitude & Area)
A right triangle with perpendicular legs $a$ and $b$ has hypotenuse $c = \sqrt{a^2 + b^2}$, complementary acute angles $\alpha = \arctan(b/a)$ and $\beta = 90^\circ - \alpha$, and area $A = \frac{1}{2}ab$.
c = √(a² + b²), tan(α) = b / a, Area = ½·a·bInputs & Parameters (2)
u
1u±0.530u
u
1u±0.530u
Hypotenuse c = √(a²+b²)
15
c² = 144 + 81 = 225
Acute Angles (α, β)
36.87° / 53.13°
Grade = 75% slope
Area & Perimeter
54 u²
Perimeter P = 36 u
Altitude h_c & Inradius r
h_c = 7.2
Inradius r = 3
20-Row Comparison Table
Compare across:
| # | Adjacent Base Leg (a) (u) | Hypotenuse (c) | Angle α | Area (½ab) |
|---|---|---|---|---|
| 01 | 1 u | 9.055 u | 83.66° | 4.5 u² |
| 02 | 2.5 u | 9.341 u | 74.48° | 11.25 u² |
| 03 | 4 u | 9.849 u | 66.04° | 18 u² |
| 04 | 5.5 u | 10.548 u | 58.57° | 24.75 u² |
| 05 | 7 u | 11.402 u | 52.13° | 31.5 u² |
| 06 | 8.5 u | 12.379 u | 46.64° | 38.25 u² |
| 07 | 10 u | 13.454 u | 41.99° | 45 u² |
| 08 | 11.5 u | 14.603 u | 38.05° | 51.75 u² |
| 09 | 13 u | 15.811 u | 34.7° | 58.5 u² |
| 10 | 14.5 u | 17.066 u | 31.83° | 65.25 u² |
| 11 | 16.5 u | 18.795 u | 28.61° | 74.25 u² |
| 12 | 18 u | 20.125 u | 26.57° | 81 u² |
| 13 | 19.5 u | 21.477 u | 24.78° | 87.75 u² |
| 14 | 21 u | 22.847 u | 23.2° | 94.5 u² |
| 15 | 22.5 u | 24.233 u | 21.8° | 101.25 u² |
| 16 | 24 u | 25.632 u | 20.56° | 108 u² |
| 17 | 25.5 u | 27.042 u | 19.44° | 114.75 u² |
| 18 | 27 u | 28.46 u | 18.43° | 121.5 u² |
| 19 | 28.5 u | 29.887 u | 17.53° | 128.25 u² |
| 20 | 30 u | 31.321 u | 16.7° | 135 u² |
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Key Analytical Takeaways
- 01The internal altitude $h_c = \frac{ab}{c}$ dropped to the hypotenuse divides the triangle into two smaller similar right triangles.
- 02Classic integer Pythagorean triples such as $(3,4,5)$, $(5,12,13)$, $(8,15,17)$, and $(12,9,15)$ yield exact integer hypotenuses.
- 03The circumcircle radius of any right triangle is exactly half the hypotenuse ($R = c/2$), while the incircle radius is $r = (a + b - c)/2$.
Frequently Asked Questions
What does SOHCAHTOA stand for in right-triangle trigonometry?+
Sine = Opposite / Hypotenuse, Cosine = Adjacent / Hypotenuse, and Tangent = Opposite / Adjacent.
How is the altitude to the hypotenuse calculated?+
By equating area formulas ½·a·b = ½·c·h_c, the perpendicular altitude to the hypotenuse is always h_c = (a × b) / c.
How is the altitude to the hypotenuse calculated in a right triangle?+
Dropping a perpendicular altitude h from the right angle to hypotenuse c divides the triangle into similar sub-triangles, giving h = (a × b) / c.