Triangle calculator

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

You have entered side a, b and angle α.

Triangle has two solutions: a=54; b=62; c=11.05664267867 and a=54; b=62; c=83.933308416.

#1 Obtuse scalene triangle.

Sides: a = 54   b = 62   c = 11.05664267867

Area: T = 220.3154958523
Perimeter: p = 127.0566426787
Semiperimeter: s = 63.52882133934

Angle ∠ A = α = 40° = 0.69881317008 rad
Angle ∠ B = β = 132.4377368296° = 132°26'15″ = 2.31114681294 rad
Angle ∠ C = γ = 7.56326317037° = 7°33'45″ = 0.13219928233 rad

Height: ha = 8.16598132786
Height: hb = 7.10769341459
Height: hc = 39.85328318006

Median: ma = 35.41435890111
Median: mb = 23.62546118835
Median: mc = 57.8744336771

Inradius: r = 3.46879860609
Circumradius: R = 42.00545433252

Vertex coordinates: A[11.05664267867; 0] B[0; 0] C[-36.43883286866; 39.85328318006]
Centroid: CG[-8.46106339666; 13.28442772669]
Coordinates of the circumscribed circle: U[5.52882133934; 41.63991704605]
Coordinates of the inscribed circle: I[1.52882133934; 3.46879860609]

Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 140° = 0.69881317008 rad
∠ B' = β' = 47.56326317037° = 47°33'45″ = 2.31114681294 rad
∠ C' = γ' = 172.4377368296° = 172°26'15″ = 0.13219928233 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 a, b and angle α.

a=54 b=62 α=40a = 54 \ \\ b = 62 \ \\ α = 40^\circ

2. From angle α, side b and side a we calculate side c - by using the law of cosines and quadratic equation:

a2=b2+c22bccosα  542=622+c22 62 c cos40   c294.99c+928=0  p=1;q=94.99;r=928 D=q24pr=94.99241928=5311.0071899 D>0  c1,2=q±D2p=94.99±5311.012 c1,2=47.49475547±36.4383286866 c1=83.93308416 c2=11.0564267867   Factored form of the equation:  (c83.93308416)(c11.0564267867)=0   c>0  c=83.933a^2 = b^2 + c^2 - 2b c \cos α \ \\ \ \\ 54^2 = 62^2 + c^2 - 2 \cdot \ 62 \cdot \ c \cdot \ \cos 40^\circ \ \\ \ \\ \ \\ c^2 -94.99c +928 =0 \ \\ \ \\ p=1; q=-94.99; r=928 \ \\ D = q^2 - 4pr = 94.99^2 - 4\cdot 1 \cdot 928 = 5311.0071899 \ \\ D>0 \ \\ \ \\ c_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 94.99 \pm \sqrt{ 5311.01 } }{ 2 } \ \\ c_{1,2} = 47.49475547 \pm 36.4383286866 \ \\ c_{1} = 83.93308416 \ \\ c_{2} = 11.0564267867 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (c -83.93308416) (c -11.0564267867) = 0 \ \\ \ \\ \ \\ c > 0 \ \\ \ \\ c = 83.933

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=54 b=62 c=11.06a = 54 \ \\ b = 62 \ \\ c = 11.06

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

p=a+b+c=54+62+11.06=127.06p = a+b+c = 54+62+11.06 = 127.06

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=127.062=63.53s = \dfrac{ p }{ 2 } = \dfrac{ 127.06 }{ 2 } = 63.53

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=63.53(63.5354)(63.5362)(63.5311.06) T=48538.68=220.31T = \sqrt{ s(s-a)(s-b)(s-c) } \ \\ T = \sqrt{ 63.53(63.53-54)(63.53-62)(63.53-11.06) } \ \\ T = \sqrt{ 48538.68 } = 220.31

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=2 220.3154=8.16 hb=2 Tb=2 220.3162=7.11 hc=2 Tc=2 220.3111.06=39.85T = \dfrac{ a h _a }{ 2 } \ \\ \ \\ h _a = \dfrac{ 2 \ T }{ a } = \dfrac{ 2 \cdot \ 220.31 }{ 54 } = 8.16 \ \\ h _b = \dfrac{ 2 \ T }{ b } = \dfrac{ 2 \cdot \ 220.31 }{ 62 } = 7.11 \ \\ h _c = \dfrac{ 2 \ T }{ c } = \dfrac{ 2 \cdot \ 220.31 }{ 11.06 } = 39.85

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(622+11.0625422 62 11.06)=40  b2=a2+c22accosβ β=arccos(a2+c2b22ac)=arccos(542+11.0626222 54 11.06)=1322615" γ=180αβ=180401322615"=73345"a^2 = b^2+c^2 - 2bc \cos α \ \\ \ \\ α = \arccos(\dfrac{ b^2+c^2-a^2 }{ 2bc } ) = \arccos(\dfrac{ 62^2+11.06^2-54^2 }{ 2 \cdot \ 62 \cdot \ 11.06 } ) = 40^\circ \ \\ \ \\ b^2 = a^2+c^2 - 2ac \cos β \ \\ β = \arccos(\dfrac{ a^2+c^2-b^2 }{ 2ac } ) = \arccos(\dfrac{ 54^2+11.06^2-62^2 }{ 2 \cdot \ 54 \cdot \ 11.06 } ) = 132^\circ 26'15" \ \\ γ = 180^\circ - α - β = 180^\circ - 40^\circ - 132^\circ 26'15" = 7^\circ 33'45"

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=220.3163.53=3.47T = rs \ \\ r = \dfrac{ T }{ s } = \dfrac{ 220.31 }{ 63.53 } = 3.47

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=54 62 11.064 3.468 63.528=42R = \dfrac{ a b c }{ 4 \ r s } = \dfrac{ 54 \cdot \ 62 \cdot \ 11.06 }{ 4 \cdot \ 3.468 \cdot \ 63.528 } = 42

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=2 622+2 11.0625422=35.414 mb=2c2+2a2b22=2 11.062+2 5426222=23.625 mc=2a2+2b2c22=2 542+2 62211.0622=57.874m_a = \dfrac{ \sqrt{ 2b^2+2c^2 - a^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 62^2+2 \cdot \ 11.06^2 - 54^2 } }{ 2 } = 35.414 \ \\ m_b = \dfrac{ \sqrt{ 2c^2+2a^2 - b^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 11.06^2+2 \cdot \ 54^2 - 62^2 } }{ 2 } = 23.625 \ \\ m_c = \dfrac{ \sqrt{ 2a^2+2b^2 - c^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 54^2+2 \cdot \ 62^2 - 11.06^2 } }{ 2 } = 57.874



#2 Obtuse scalene triangle.

Sides: a = 54   b = 62   c = 83.933308416

Area: T = 1672.486554277
Perimeter: p = 199.933308416
Semiperimeter: s = 99.967654208

Angle ∠ A = α = 40° = 0.69881317008 rad
Angle ∠ B = β = 47.56326317037° = 47°33'45″ = 0.83301245241 rad
Angle ∠ C = γ = 92.43773682963° = 92°26'15″ = 1.61333364286 rad

Height: ha = 61.94439089913
Height: hb = 53.95111465408
Height: hc = 39.85328318006

Median: ma = 68.66986340938
Median: mb = 63.39985907439
Median: mc = 40.23444298561

Inradius: r = 16.73304530893
Circumradius: R = 42.00545433252

Vertex coordinates: A[83.933308416; 0] B[0; 0] C[36.43883286866; 39.85328318006]
Centroid: CG[40.12438042822; 13.28442772669]
Coordinates of the circumscribed circle: U[41.967654208; -1.78663386599]
Coordinates of the inscribed circle: I[37.967654208; 16.73304530893]

Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 140° = 0.69881317008 rad
∠ B' = β' = 132.4377368296° = 132°26'15″ = 0.83301245241 rad
∠ C' = γ' = 87.56326317037° = 87°33'45″ = 1.61333364286 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 a, b and angle α.

a=54 b=62 α=40a = 54 \ \\ b = 62 \ \\ α = 40^\circ

2. From angle α, side b and side a we calculate side c - by using the law of cosines and quadratic equation:

a2=b2+c22bccosα  542=622+c22 62 c cos40   c294.99c+928=0  p=1;q=94.99;r=928 D=q24pr=94.99241928=5311.0071899 D>0  c1,2=q±D2p=94.99±5311.012 c1,2=47.49475547±36.4383286866 c1=83.93308416 c2=11.0564267867   Factored form of the equation:  (c83.93308416)(c11.0564267867)=0   c>0  c=83.933a^2 = b^2 + c^2 - 2b c \cos α \ \\ \ \\ 54^2 = 62^2 + c^2 - 2 \cdot \ 62 \cdot \ c \cdot \ \cos 40^\circ \ \\ \ \\ \ \\ c^2 -94.99c +928 =0 \ \\ \ \\ p=1; q=-94.99; r=928 \ \\ D = q^2 - 4pr = 94.99^2 - 4\cdot 1 \cdot 928 = 5311.0071899 \ \\ D>0 \ \\ \ \\ c_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 94.99 \pm \sqrt{ 5311.01 } }{ 2 } \ \\ c_{1,2} = 47.49475547 \pm 36.4383286866 \ \\ c_{1} = 83.93308416 \ \\ c_{2} = 11.0564267867 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (c -83.93308416) (c -11.0564267867) = 0 \ \\ \ \\ \ \\ c > 0 \ \\ \ \\ c = 83.933

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=54 b=62 c=83.93a = 54 \ \\ b = 62 \ \\ c = 83.93

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

p=a+b+c=54+62+83.93=199.93p = a+b+c = 54+62+83.93 = 199.93

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=199.932=99.97s = \dfrac{ p }{ 2 } = \dfrac{ 199.93 }{ 2 } = 99.97

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=99.97(99.9754)(99.9762)(99.9783.93) T=2797207.89=1672.49T = \sqrt{ s(s-a)(s-b)(s-c) } \ \\ T = \sqrt{ 99.97(99.97-54)(99.97-62)(99.97-83.93) } \ \\ T = \sqrt{ 2797207.89 } = 1672.49

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 1672.4954=61.94 hb=2 Tb=2 1672.4962=53.95 hc=2 Tc=2 1672.4983.93=39.85T = \dfrac{ a h _a }{ 2 } \ \\ \ \\ h _a = \dfrac{ 2 \ T }{ a } = \dfrac{ 2 \cdot \ 1672.49 }{ 54 } = 61.94 \ \\ h _b = \dfrac{ 2 \ T }{ b } = \dfrac{ 2 \cdot \ 1672.49 }{ 62 } = 53.95 \ \\ h _c = \dfrac{ 2 \ T }{ c } = \dfrac{ 2 \cdot \ 1672.49 }{ 83.93 } = 39.85

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(622+83.9325422 62 83.93)=40  b2=a2+c22accosβ β=arccos(a2+c2b22ac)=arccos(542+83.9326222 54 83.93)=473345" γ=180αβ=18040473345"=922615"a^2 = b^2+c^2 - 2bc \cos α \ \\ \ \\ α = \arccos(\dfrac{ b^2+c^2-a^2 }{ 2bc } ) = \arccos(\dfrac{ 62^2+83.93^2-54^2 }{ 2 \cdot \ 62 \cdot \ 83.93 } ) = 40^\circ \ \\ \ \\ b^2 = a^2+c^2 - 2ac \cos β \ \\ β = \arccos(\dfrac{ a^2+c^2-b^2 }{ 2ac } ) = \arccos(\dfrac{ 54^2+83.93^2-62^2 }{ 2 \cdot \ 54 \cdot \ 83.93 } ) = 47^\circ 33'45" \ \\ γ = 180^\circ - α - β = 180^\circ - 40^\circ - 47^\circ 33'45" = 92^\circ 26'15"

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=1672.4999.97=16.73T = rs \ \\ r = \dfrac{ T }{ s } = \dfrac{ 1672.49 }{ 99.97 } = 16.73

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=54 62 83.934 16.73 99.967=42R = \dfrac{ a b c }{ 4 \ r s } = \dfrac{ 54 \cdot \ 62 \cdot \ 83.93 }{ 4 \cdot \ 16.73 \cdot \ 99.967 } = 42

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=2 622+2 83.9325422=68.669 mb=2c2+2a2b22=2 83.932+2 5426222=63.399 mc=2a2+2b2c22=2 542+2 62283.9322=40.234m_a = \dfrac{ \sqrt{ 2b^2+2c^2 - a^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 62^2+2 \cdot \ 83.93^2 - 54^2 } }{ 2 } = 68.669 \ \\ m_b = \dfrac{ \sqrt{ 2c^2+2a^2 - b^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 83.93^2+2 \cdot \ 54^2 - 62^2 } }{ 2 } = 63.399 \ \\ m_c = \dfrac{ \sqrt{ 2a^2+2b^2 - c^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 54^2+2 \cdot \ 62^2 - 83.93^2 } }{ 2 } = 40.234

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