Triangle calculator

Please enter what you know about the triangle:
Symbols definition of ABC triangle

You have entered side b, c and area T.

Triangle has two solutions: a=31.11774184616; b=21.9; c=10.4 and a=14.39660504404; b=21.9; c=10.4.

#1 Obtuse scalene triangle.

Sides: a = 31.11774184616   b = 21.9   c = 10.4

Area: T = 62.6
Perimeter: p = 63.41774184616
Semiperimeter: s = 31.70987092308

Angle ∠ A = α = 146.6533467075° = 146°39'12″ = 2.56595858599 rad
Angle ∠ B = β = 22.76600128159° = 22°45'36″ = 0.39772371614 rad
Angle ∠ C = γ = 10.58765201095° = 10°35'11″ = 0.18547696322 rad

Height: ha = 4.0233470011
Height: hb = 5.71768949772
Height: hc = 12.03884615385

Median: ma = 7.19880252202
Median: mb = 20.45329793883
Median: mc = 26.39990883528

Inradius: r = 1.97442210112
Circumradius: R = 28.30439266327

Vertex coordinates: A[10.4; 0] B[0; 0] C[28.69444101787; 12.03884615385]
Centroid: CG[13.03114700596; 4.01328205128]
Coordinates of the circumscribed circle: U[5.2; 27.8222154173]
Coordinates of the inscribed circle: I[9.80987092308; 1.97442210112]

Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 33.34765329254° = 33°20'48″ = 2.56595858599 rad
∠ B' = β' = 157.2439987184° = 157°14'24″ = 0.39772371614 rad
∠ C' = γ' = 169.413347989° = 169°24'49″ = 0.18547696322 rad


How did we calculate this triangle?

The calculation of the triangle progress in two phases. The first phase is such that we try to calculate all three sides of the triangle from the input parameters. The first phase is different for the different triangles query entered. The second phase is the calculation of other characteristics of the triangle, such as angles, area, perimeter, heights, the center of gravity, circle radii, etc. Some input data also results in two to three correct triangle solutions (e.g., if the specified triangle area and two sides - typically resulting in both acute and obtuse) triangle).

1. Input data entered: side b, c and area T.

b=21.9 c=10.4 T=62.6b = 21.9 \ \\ c = 10.4 \ \\ T = 62.6

2. From area T, side b and side c we calculate side a - using Heron's formula for the area and solve of the bikvadratic equation:

s=a+b+c2 T2=s(sa)(sb)(sc)  s=a+21.9+10.42=a+32.32=a/2+16.15  T2=s(sa)(sb)(sc) T2=(a/2+16.15)(a/2+16.15a)(a/2+16.1521.9)(a/2+16.1510.4)  62.62=(a/2+16.15)(16.15a/2)(a/2+(5.75))(a/2+5.75) 62700.16=(a+32.3)(32.3a)(a+(11.5))(a+11.5)  D=b2 c24 T2=21.92 10.424 62.62=36199.578  D1=2 D+b2+c2=2 36199.578+21.92+10.42=207.246 D2=2 D+b2+c2=2 36199.578+21.92+10.42=968.294  a1=D1=207.246=14.396 a2=D2=968.294=31.117s = \dfrac{ a+b+c }{ 2 } \ \\ T^2 = s(s-a)(s-b)(s-c) \ \\ \ \\ s = \dfrac{ a+21.9+10.4 }{ 2 } = \dfrac{ a+32.3 }{ 2 } = a/2 + 16.15 \ \\ \ \\ T^2 = s(s-a)(s-b)(s-c) \ \\ T^2 = ( a/2 + 16.15) ( a/2 + 16.15-a) ( a/2 + 16.15-21.9) ( a/2 + 16.15 - 10.4) \ \\ \ \\ 62.6^2 = ( a/2 + 16.15) ( 16.15-a/2) ( a/2 + (-5.75)) ( a/2 + 5.75) \ \\ 62700.16 = ( a + 32.3) ( 32.3-a) ( a + (-11.5)) ( a + 11.5) \ \\ \ \\ D = b^2 \cdot \ c^2 - 4 \cdot \ T^2 = 21.9^2 \cdot \ 10.4^2 - 4 \cdot \ 62.6^2 = 36199.578 \ \\ \ \\ D_1 = -2 \cdot \ \sqrt{ D } + b^2 + c^2 = -2 \cdot \ \sqrt{ 36199.578 } + 21.9^2 + 10.4^2 = 207.246 \ \\ D_2 = 2 \cdot \ \sqrt{ D } + b^2 + c^2 = 2 \cdot \ \sqrt{ 36199.578 } + 21.9^2 + 10.4^2 = 968.294 \ \\ \ \\ a_1 = \sqrt{ D_1 } = \sqrt{ 207.246 } = 14.396 \ \\ a_2 = \sqrt{ D_2 } = \sqrt{ 968.294 } = 31.117

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=31.12 b=21.9 c=10.4a = 31.12 \ \\ b = 21.9 \ \\ c = 10.4

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

p=a+b+c=31.12+21.9+10.4=63.42p = a+b+c = 31.12+21.9+10.4 = 63.42

4. Semiperimeter of the triangle

The semiperimeter of the triangle is half its perimeter. The semiperimeter frequently appears in formulas for triangles that it is given a separate name. By the triangle inequality, the longest side length of a triangle is less than the semiperimeter.

s=p2=63.422=31.71s = \dfrac{ p }{ 2 } = \dfrac{ 63.42 }{ 2 } = 31.71

5. The triangle area using Heron's formula

Heron's formula gives the area of a triangle when the length of all three sides are known. There is no need to calculate angles or other distances in the triangle first. Heron's formula works equally well in all cases and types of triangles.

T=s(sa)(sb)(sc) T=31.71(31.7131.12)(31.7121.9)(31.7110.4) T=3918.76=62.6T = \sqrt{ s(s-a)(s-b)(s-c) } \ \\ T = \sqrt{ 31.71(31.71-31.12)(31.71-21.9)(31.71-10.4) } \ \\ T = \sqrt{ 3918.76 } = 62.6

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

There are many ways to find the height of the triangle. The easiest way is from the area and base length. The area of a triangle is half of the product of the length of the base and the height. Every side of the triangle can be a base; there are three bases and three heights (altitudes). Triangle height is the perpendicular line segment from a vertex to a line containing the base.

T=aha2  ha=2 Ta=2 62.631.12=4.02 hb=2 Tb=2 62.621.9=5.72 hc=2 Tc=2 62.610.4=12.04T = \dfrac{ a h _a }{ 2 } \ \\ \ \\ h _a = \dfrac{ 2 \ T }{ a } = \dfrac{ 2 \cdot \ 62.6 }{ 31.12 } = 4.02 \ \\ h _b = \dfrac{ 2 \ T }{ b } = \dfrac{ 2 \cdot \ 62.6 }{ 21.9 } = 5.72 \ \\ h _c = \dfrac{ 2 \ T }{ c } = \dfrac{ 2 \cdot \ 62.6 }{ 10.4 } = 12.04

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

The Law of Cosines is useful for finding the angles of a triangle when we know all three sides. The cosine rule, also known as the law of cosines, relates all three sides of a triangle with an angle of a triangle. The Law of Cosines is the extrapolation of the Pythagorean theorem for any triangle. Pythagorean theorem works only in a right triangle. Pythagorean theorem is a special case of the Law of Cosines and can be derived from it because the cosine of 90° is 0. It is best to find the angle opposite the longest side first. With the Law of Cosines, there is also no problem with obtuse angles as with the Law of Sines, because cosine function is negative for obtuse angles, zero for right, and positive for acute angles. We also use inverse cosine called arccosine to determine the angle from cosine value.

a2=b2+c22bccosα  α=arccos(b2+c2a22bc)=arccos(21.92+10.4231.1222 21.9 10.4)=1463912"  b2=a2+c22accosβ β=arccos(a2+c2b22ac)=arccos(31.122+10.4221.922 31.12 10.4)=224536" γ=180αβ=1801463912"224536"=103511"a^2 = b^2+c^2 - 2bc \cos α \ \\ \ \\ α = \arccos(\dfrac{ b^2+c^2-a^2 }{ 2bc } ) = \arccos(\dfrac{ 21.9^2+10.4^2-31.12^2 }{ 2 \cdot \ 21.9 \cdot \ 10.4 } ) = 146^\circ 39'12" \ \\ \ \\ b^2 = a^2+c^2 - 2ac \cos β \ \\ β = \arccos(\dfrac{ a^2+c^2-b^2 }{ 2ac } ) = \arccos(\dfrac{ 31.12^2+10.4^2-21.9^2 }{ 2 \cdot \ 31.12 \cdot \ 10.4 } ) = 22^\circ 45'36" \ \\ γ = 180^\circ - α - β = 180^\circ - 146^\circ 39'12" - 22^\circ 45'36" = 10^\circ 35'11"

8. Inradius

An incircle of a triangle is a circle which is tangent to each side. An incircle center is called incenter and has a radius named inradius. All triangles have an incenter, and it always lies inside the triangle. The incenter is the intersection of the three angle bisectors. The product of the inradius and semiperimeter (half the perimeter) of a triangle is its area.

T=rs r=Ts=62.631.71=1.97T = rs \ \\ r = \dfrac{ T }{ s } = \dfrac{ 62.6 }{ 31.71 } = 1.97

9. Circumradius

The circumcircle of a triangle is a circle that passes through all of the triangle's vertices, and the circumradius of a triangle is the radius of the triangle's circumcircle. Circumcenter (center of circumcircle) is the point where the perpendicular bisectors of a triangle intersect.

R=abc4 rs=31.12 21.9 10.44 1.974 31.709=28.3R = \dfrac{ a b c }{ 4 \ r s } = \dfrac{ 31.12 \cdot \ 21.9 \cdot \ 10.4 }{ 4 \cdot \ 1.974 \cdot \ 31.709 } = 28.3

10. Calculation of medians

A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. Every triangle has three medians, and they all intersect each other at the triangle's centroid. The centroid divides each median into parts in the ratio 2:1, with the centroid being twice as close to the midpoint of a side as it is to the opposite vertex. We use Apollonius's theorem to calculate the length of a median from the lengths of its side.

ma=2b2+2c2a22=2 21.92+2 10.4231.1222=7.198 mb=2c2+2a2b22=2 10.42+2 31.12221.922=20.453 mc=2a2+2b2c22=2 31.122+2 21.9210.422=26.399m_a = \dfrac{ \sqrt{ 2b^2+2c^2 - a^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 21.9^2+2 \cdot \ 10.4^2 - 31.12^2 } }{ 2 } = 7.198 \ \\ m_b = \dfrac{ \sqrt{ 2c^2+2a^2 - b^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 10.4^2+2 \cdot \ 31.12^2 - 21.9^2 } }{ 2 } = 20.453 \ \\ m_c = \dfrac{ \sqrt{ 2a^2+2b^2 - c^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 31.12^2+2 \cdot \ 21.9^2 - 10.4^2 } }{ 2 } = 26.399



#2 Obtuse scalene triangle.

Sides: a = 14.39660504404   b = 21.9   c = 10.4

Area: T = 62.6
Perimeter: p = 46.69660504404
Semiperimeter: s = 23.34880252202

Angle ∠ A = α = 33.34765329254° = 33°20'48″ = 0.58220067937 rad
Angle ∠ B = β = 123.2555488315° = 123°15'20″ = 2.15112140922 rad
Angle ∠ C = γ = 23.39879787596° = 23°23'53″ = 0.40883717677 rad

Height: ha = 8.69768297672
Height: hb = 5.71768949772
Height: hc = 12.03884615385

Median: ma = 15.55987092308
Median: mb = 6.14882220309
Median: mc = 17.787730261

Inradius: r = 2.68111689387
Circumradius: R = 13.09444267105

Vertex coordinates: A[10.4; 0] B[0; 0] C[-7.89444101787; 12.03884615385]
Centroid: CG[0.83551966071; 4.01328205128]
Coordinates of the circumscribed circle: U[5.2; 12.01876541337]
Coordinates of the inscribed circle: I[1.44880252202; 2.68111689387]

Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 146.6533467075° = 146°39'12″ = 0.58220067937 rad
∠ B' = β' = 56.74545116851° = 56°44'40″ = 2.15112140922 rad
∠ C' = γ' = 156.602202124° = 156°36'7″ = 0.40883717677 rad

Calculate another triangle

How did we calculate this triangle?

The calculation of the triangle progress in two phases. The first phase is such that we try to calculate all three sides of the triangle from the input parameters. The first phase is different for the different triangles query entered. The second phase is the calculation of other characteristics of the triangle, such as angles, area, perimeter, heights, the center of gravity, circle radii, etc. Some input data also results in two to three correct triangle solutions (e.g., if the specified triangle area and two sides - typically resulting in both acute and obtuse) triangle).

1. Input data entered: side b, c and area T.

b=21.9 c=10.4 T=62.6b = 21.9 \ \\ c = 10.4 \ \\ T = 62.6

2. From area T, side b and side c we calculate side a - using Heron's formula for the area and solve of the bikvadratic equation:

s=a+b+c2 T2=s(sa)(sb)(sc)  s=a+21.9+10.42=a+32.32=a/2+16.15  T2=s(sa)(sb)(sc) T2=(a/2+16.15)(a/2+16.15a)(a/2+16.1521.9)(a/2+16.1510.4)  62.62=(a/2+16.15)(16.15a/2)(a/2+(5.75))(a/2+5.75) 62700.16=(a+32.3)(32.3a)(a+(11.5))(a+11.5)  D=b2 c24 T2=21.92 10.424 62.62=36199.578  D1=2 D+b2+c2=2 36199.578+21.92+10.42=207.246 D2=2 D+b2+c2=2 36199.578+21.92+10.42=968.294  a1=D1=207.246=14.396 a2=D2=968.294=31.117s = \dfrac{ a+b+c }{ 2 } \ \\ T^2 = s(s-a)(s-b)(s-c) \ \\ \ \\ s = \dfrac{ a+21.9+10.4 }{ 2 } = \dfrac{ a+32.3 }{ 2 } = a/2 + 16.15 \ \\ \ \\ T^2 = s(s-a)(s-b)(s-c) \ \\ T^2 = ( a/2 + 16.15) ( a/2 + 16.15-a) ( a/2 + 16.15-21.9) ( a/2 + 16.15 - 10.4) \ \\ \ \\ 62.6^2 = ( a/2 + 16.15) ( 16.15-a/2) ( a/2 + (-5.75)) ( a/2 + 5.75) \ \\ 62700.16 = ( a + 32.3) ( 32.3-a) ( a + (-11.5)) ( a + 11.5) \ \\ \ \\ D = b^2 \cdot \ c^2 - 4 \cdot \ T^2 = 21.9^2 \cdot \ 10.4^2 - 4 \cdot \ 62.6^2 = 36199.578 \ \\ \ \\ D_1 = -2 \cdot \ \sqrt{ D } + b^2 + c^2 = -2 \cdot \ \sqrt{ 36199.578 } + 21.9^2 + 10.4^2 = 207.246 \ \\ D_2 = 2 \cdot \ \sqrt{ D } + b^2 + c^2 = 2 \cdot \ \sqrt{ 36199.578 } + 21.9^2 + 10.4^2 = 968.294 \ \\ \ \\ a_1 = \sqrt{ D_1 } = \sqrt{ 207.246 } = 14.396 \ \\ a_2 = \sqrt{ D_2 } = \sqrt{ 968.294 } = 31.117

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=14.4 b=21.9 c=10.4a = 14.4 \ \\ b = 21.9 \ \\ c = 10.4

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

p=a+b+c=14.4+21.9+10.4=46.7p = a+b+c = 14.4+21.9+10.4 = 46.7

4. Semiperimeter of the triangle

The semiperimeter of the triangle is half its perimeter. The semiperimeter frequently appears in formulas for triangles that it is given a separate name. By the triangle inequality, the longest side length of a triangle is less than the semiperimeter.

s=p2=46.72=23.35s = \dfrac{ p }{ 2 } = \dfrac{ 46.7 }{ 2 } = 23.35

5. The triangle area using Heron's formula

Heron's formula gives the area of a triangle when the length of all three sides are known. There is no need to calculate angles or other distances in the triangle first. Heron's formula works equally well in all cases and types of triangles.

T=s(sa)(sb)(sc) T=23.35(23.3514.4)(23.3521.9)(23.3510.4) T=3918.76=62.6T = \sqrt{ s(s-a)(s-b)(s-c) } \ \\ T = \sqrt{ 23.35(23.35-14.4)(23.35-21.9)(23.35-10.4) } \ \\ T = \sqrt{ 3918.76 } = 62.6

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

There are many ways to find the height of the triangle. The easiest way is from the area and base length. The area of a triangle is half of the product of the length of the base and the height. Every side of the triangle can be a base; there are three bases and three heights (altitudes). Triangle height is the perpendicular line segment from a vertex to a line containing the base.

T=aha2  ha=2 Ta=2 62.614.4=8.7 hb=2 Tb=2 62.621.9=5.72 hc=2 Tc=2 62.610.4=12.04T = \dfrac{ a h _a }{ 2 } \ \\ \ \\ h _a = \dfrac{ 2 \ T }{ a } = \dfrac{ 2 \cdot \ 62.6 }{ 14.4 } = 8.7 \ \\ h _b = \dfrac{ 2 \ T }{ b } = \dfrac{ 2 \cdot \ 62.6 }{ 21.9 } = 5.72 \ \\ h _c = \dfrac{ 2 \ T }{ c } = \dfrac{ 2 \cdot \ 62.6 }{ 10.4 } = 12.04

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

The Law of Cosines is useful for finding the angles of a triangle when we know all three sides. The cosine rule, also known as the law of cosines, relates all three sides of a triangle with an angle of a triangle. The Law of Cosines is the extrapolation of the Pythagorean theorem for any triangle. Pythagorean theorem works only in a right triangle. Pythagorean theorem is a special case of the Law of Cosines and can be derived from it because the cosine of 90° is 0. It is best to find the angle opposite the longest side first. With the Law of Cosines, there is also no problem with obtuse angles as with the Law of Sines, because cosine function is negative for obtuse angles, zero for right, and positive for acute angles. We also use inverse cosine called arccosine to determine the angle from cosine value.

a2=b2+c22bccosα  α=arccos(b2+c2a22bc)=arccos(21.92+10.4214.422 21.9 10.4)=332048"  b2=a2+c22accosβ β=arccos(a2+c2b22ac)=arccos(14.42+10.4221.922 14.4 10.4)=1231520" γ=180αβ=180332048"1231520"=232353"a^2 = b^2+c^2 - 2bc \cos α \ \\ \ \\ α = \arccos(\dfrac{ b^2+c^2-a^2 }{ 2bc } ) = \arccos(\dfrac{ 21.9^2+10.4^2-14.4^2 }{ 2 \cdot \ 21.9 \cdot \ 10.4 } ) = 33^\circ 20'48" \ \\ \ \\ b^2 = a^2+c^2 - 2ac \cos β \ \\ β = \arccos(\dfrac{ a^2+c^2-b^2 }{ 2ac } ) = \arccos(\dfrac{ 14.4^2+10.4^2-21.9^2 }{ 2 \cdot \ 14.4 \cdot \ 10.4 } ) = 123^\circ 15'20" \ \\ γ = 180^\circ - α - β = 180^\circ - 33^\circ 20'48" - 123^\circ 15'20" = 23^\circ 23'53"

8. Inradius

An incircle of a triangle is a circle which is tangent to each side. An incircle center is called incenter and has a radius named inradius. All triangles have an incenter, and it always lies inside the triangle. The incenter is the intersection of the three angle bisectors. The product of the inradius and semiperimeter (half the perimeter) of a triangle is its area.

T=rs r=Ts=62.623.35=2.68T = rs \ \\ r = \dfrac{ T }{ s } = \dfrac{ 62.6 }{ 23.35 } = 2.68

9. Circumradius

The circumcircle of a triangle is a circle that passes through all of the triangle's vertices, and the circumradius of a triangle is the radius of the triangle's circumcircle. Circumcenter (center of circumcircle) is the point where the perpendicular bisectors of a triangle intersect.

R=abc4 rs=14.4 21.9 10.44 2.681 23.348=13.09R = \dfrac{ a b c }{ 4 \ r s } = \dfrac{ 14.4 \cdot \ 21.9 \cdot \ 10.4 }{ 4 \cdot \ 2.681 \cdot \ 23.348 } = 13.09

10. Calculation of medians

A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. Every triangle has three medians, and they all intersect each other at the triangle's centroid. The centroid divides each median into parts in the ratio 2:1, with the centroid being twice as close to the midpoint of a side as it is to the opposite vertex. We use Apollonius's theorem to calculate the length of a median from the lengths of its side.

ma=2b2+2c2a22=2 21.92+2 10.4214.422=15.559 mb=2c2+2a2b22=2 10.42+2 14.4221.922=6.148 mc=2a2+2b2c22=2 14.42+2 21.9210.422=17.787m_a = \dfrac{ \sqrt{ 2b^2+2c^2 - a^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 21.9^2+2 \cdot \ 10.4^2 - 14.4^2 } }{ 2 } = 15.559 \ \\ m_b = \dfrac{ \sqrt{ 2c^2+2a^2 - b^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 10.4^2+2 \cdot \ 14.4^2 - 21.9^2 } }{ 2 } = 6.148 \ \\ m_c = \dfrac{ \sqrt{ 2a^2+2b^2 - c^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 14.4^2+2 \cdot \ 21.9^2 - 10.4^2 } }{ 2 } = 17.787

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