Triangle calculator

You have entered side b, c and height h_c.

Acute scalene triangle.

Sides: a = 52.15440046522 b = 48 c = 60

Area: T = 1200
Perimeter: p = 160.1544004652
Semiperimeter: s = 80.07770023261

Angle ∠ A = α = 56.44326902381° = 56°26'34″ = 0.98551107833 rad
Angle ∠ B = β = 50.08216182447° = 50°4'54″ = 0.87440891331 rad
Angle ∠ C = γ = 73.47656915172° = 73°28'32″ = 1.28223927372 rad

Height: h_a = 46.01875592652
Height: h_b = 50
Height: h_c = 40

Median: m_a = 47.66553957257
Median: m_b = 50.83332578203
Median: m_c = 40.15499701199

Inradius: r = 14.98655759474
Circumradius: R = 31.29224027913

Vertex coordinates: A[60; 0] B[0; 0] C[33.46770016772; 40]
Centroid: CG[31.15656672257; 13.33333333333]
Coordinates of the circumscribed circle: U[30; 8.99002512579]
Coordinates of the inscribed circle: I[32.07770023261; 14.98655759474]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 123.5577309762° = 123°33'26″ = 0.98551107833 rad
∠ B' = β' = 129.9188381755° = 129°55'6″ = 0.87440891331 rad
∠ C' = γ' = 106.5244308483° = 106°31'28″ = 1.28223927372 rad

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How did we calculate this triangle?

1. Input data entered: side b, c and height h_c.

$b = 48 ; ; c = 60 ; ; h_c = 40 ; ;$

2. From side c we calculate T:

$T = fraction{ c h_c }{ 2 } ; ; ; ; T = fraction{ 60 * 40 }{ 2 } = 1200 ; ;$

3. From side c and angle α we calculate height h_b:

$h_b = c * sin alpha = 60 * sin 56° 26'34" = 50 ; ;$

4. Calculation of the third side a of the triangle using a Law of Cosines

$a**2 = b**2+c**2 - 2bc cos alpha ; ; a = sqrt{ b**2+c**2 - 2bc cos alpha } ; ; a = sqrt{ 48**2+60**2 - 2 * 48 * 60 * cos 56° 26'34" } ; ; a = 52.15 ; ;$
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.

$a = 52.15 ; ; b = 48 ; ; c = 60 ; ;$

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

$p = a+b+c = 52.15+48+60 = 160.15 ; ;$

6. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 160.15 }{ 2 } = 80.08 ; ;$

7. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 80.08 * (80.08-52.15)(80.08-48)(80.08-60) } ; ; T = sqrt{ 1440000 } = 1200 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 1200 }{ 52.15 } = 46.02 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 1200 }{ 48 } = 50 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 1200 }{ 60 } = 40 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos alpha ; ; alpha = arccos( fraction{ b**2+c**2-a**2 }{ 2bc } ) = arccos( fraction{ 48**2+60**2-52.15**2 }{ 2 * 48 * 60 } ) = 56° 26'34" ; ; b**2 = a**2+c**2 - 2ac cos beta ; ; beta = arccos( fraction{ a**2+c**2-b**2 }{ 2ac } ) = arccos( fraction{ 52.15**2+60**2-48**2 }{ 2 * 52.15 * 60 } ) = 50° 4'54" ; ; gamma = 180° - alpha - beta = 180° - 56° 26'34" - 50° 4'54" = 73° 28'32" ; ;$

10. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 1200 }{ 80.08 } = 14.99 ; ;$

11. Circumradius

$R = fraction{ a }{ 2 * sin alpha } = fraction{ 52.15 }{ 2 * sin 56° 26'34" } = 31.29 ; ;$

12. Calculation of medians

$m_a = fraction{ sqrt{ 2 b**2+2c**2 - a**2 } }{ 2 } = fraction{ sqrt{ 2 * 48**2+2 * 60**2 - 52.15**2 } }{ 2 } = 47.665 ; ; m_b = fraction{ sqrt{ 2 c**2+2a**2 - b**2 } }{ 2 } = fraction{ sqrt{ 2 * 60**2+2 * 52.15**2 - 48**2 } }{ 2 } = 50.833 ; ; m_c = fraction{ sqrt{ 2 b**2+2a**2 - c**2 } }{ 2 } = fraction{ sqrt{ 2 * 48**2+2 * 52.15**2 - 60**2 } }{ 2 } = 40.15 ; ;$
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