Right triangle calculator (a,b)

Please enter two properties of the right triangle

Use symbols: a, b, c, A, B, h, T, p, r, R


You have entered cathetus a and cathetus b.

Right scalene triangle.

Sides: a = 165   b = 250   c = 299.5411316015

Area: T = 20625
Perimeter: p = 714.5411316015
Semiperimeter: s = 357.2710658008

Angle ∠ A = α = 33.42548111826° = 33°25'29″ = 0.5833373007 rad
Angle ∠ B = β = 56.57551888174° = 56°34'31″ = 0.98774233198 rad
Angle ∠ C = γ = 90° = 1.57107963268 rad

Height: ha = 250
Height: hb = 165
Height: hc = 137.7110552083

Median: ma = 263.2610802248
Median: mb = 207.0022415445
Median: mc = 149.7710658008

Inradius: r = 57.72993419925
Circumradius: R = 149.7710658008

Vertex coordinates: A[299.5411316015; 0] B[0; 0] C[90.88989643746; 137.7110552083]
Centroid: CG[130.1433426796; 45.90435173609]
Coordinates of the circumscribed circle: U[149.7710658008; -0]
Coordinates of the inscribed circle: I[107.2710658008; 57.72993419925]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 146.5755188817° = 146°34'31″ = 0.5833373007 rad
∠ B' = β' = 123.4254811183° = 123°25'29″ = 0.98774233198 rad
∠ C' = γ' = 90° = 1.57107963268 rad

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How did we calculate this triangle?

1. Input data entered: cathetus a cathetus b

a = 165 ; ; b = 250 ; ;

2. From cathetus a and cathetus b we calculate hypotenuse c - Pythagorean theorem:

c**2 = a**2+b**2 ; ; c = sqrt{ a**2+b**2 } = sqrt{ 165**2 + 250**2 } = 299.541 ; ;


Now we know the lengths of all three sides of the triangle and the triangle is uniquely determined. Next we calculate another its characteristics - same procedure as calculation of the triangle from the known three sides SSS.

a = 165 ; ; b = 250 ; ; c = 299.54 ; ;

3. The triangle circumference is the sum of the lengths of its three sides

p = a+b+c = 165+250+299.54 = 714.54 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 714.54 }{ 2 } = 357.27 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 357.27 * (357.27-165)(357.27-250)(357.27-299.54) } ; ; T = sqrt{ 425390625 } = 20625 ; ;

6. Calculate the heights of the triangle from its area.

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 20625 }{ 165 } = 250 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 20625 }{ 250 } = 165 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 20625 }{ 299.54 } = 137.71 ; ;

7. Calculation of the inner angles of the triangle using a Law of Cosines

a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 165**2-250**2-299.54**2 }{ 2 * 250 * 299.54 } ) = 33° 25'29" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 250**2-165**2-299.54**2 }{ 2 * 165 * 299.54 } ) = 56° 34'31" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 299.54**2-165**2-250**2 }{ 2 * 250 * 165 } ) = 90° ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 20625 }{ 357.27 } = 57.73 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 165 }{ 2 * sin 33° 25'29" } = 149.77 ; ;
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