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 = 150   b = 1855   c = 1861.055480844

Area: T = 139125
Perimeter: p = 3866.055480844
Semiperimeter: s = 1933.027740422

Angle ∠ A = α = 4.62330231478° = 4°37'23″ = 0.08106869753 rad
Angle ∠ B = β = 85.37769768522° = 85°22'37″ = 1.49901093515 rad
Angle ∠ C = γ = 90° = 1.57107963268 rad

Height: ha = 1855
Height: hb = 150
Height: hc = 149.5121985751

Median: ma = 1856.51655534
Median: mb = 939.5511089617
Median: mc = 930.5277404218

Inradius: r = 71.97325957824
Circumradius: R = 930.5277404218

Vertex coordinates: A[1861.055480844; 0] B[0; 0] C[12.09899179852; 149.5121985751]
Centroid: CG[624.3821575473; 49.83773285836]
Coordinates of the circumscribed circle: U[930.5277404218; -0]
Coordinates of the inscribed circle: I[78.02774042176; 71.97325957824]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 175.3776976852° = 175°22'37″ = 0.08106869753 rad
∠ B' = β' = 94.62330231478° = 94°37'23″ = 1.49901093515 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 = 150 ; ; b = 1855 ; ;

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{ 150**2 + 1855**2 } = 1861.055 ; ;


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 = 150 ; ; b = 1855 ; ; c = 1861.05 ; ;

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

p = a+b+c = 150+1855+1861.05 = 3866.05 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 3866.05 }{ 2 } = 1933.03 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 1933.03 * (1933.03-150)(1933.03-1855)(1933.03-1861.05) } ; ; T = sqrt{ 19355765625 } = 139125 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 139125 }{ 150 } = 1855 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 139125 }{ 1855 } = 150 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 139125 }{ 1861.05 } = 149.51 ; ;

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{ 150**2-1855**2-1861.05**2 }{ 2 * 1855 * 1861.05 } ) = 4° 37'23" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 1855**2-150**2-1861.05**2 }{ 2 * 150 * 1861.05 } ) = 85° 22'37" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 1861.05**2-150**2-1855**2 }{ 2 * 1855 * 150 } ) = 90° ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 139125 }{ 1933.03 } = 71.97 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 150 }{ 2 * sin 4° 37'23" } = 930.53 ; ;
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