Triangle calculator SSA

Please enter two sides and a non-included angle
°


Triangle has two solutions with side c=43.94443174473 and with side c=11.62883521894

#1 Acute scalene triangle.

Sides: a = 40   b = 33   c = 43.94443174473

Area: T = 632.2187930772
Perimeter: p = 116.9444317447
Semiperimeter: s = 58.47221587237

Angle ∠ A = α = 60.68333509748° = 60°41' = 1.05991242757 rad
Angle ∠ B = β = 46° = 0.80328514559 rad
Angle ∠ C = γ = 73.31766490252° = 73°19' = 1.2879616922 rad

Height: ha = 31.61108965386
Height: hb = 38.31662382286
Height: hc = 28.77435920135

Median: ma = 33.3177435645
Median: mb = 38.64332596704
Median: mc = 29.35551399421

Inradius: r = 10.81222898927
Circumradius: R = 22.93876992518

Vertex coordinates: A[43.94443174473; 0] B[0; 0] C[27.78663348184; 28.77435920135]
Centroid: CG[23.91102174219; 9.59111973378]
Coordinates of the circumscribed circle: U[21.97221587237; 6.58550047826]
Coordinates of the inscribed circle: I[25.47221587237; 10.81222898927]

Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 119.3176649025° = 119°19' = 1.05991242757 rad
∠ B' = β' = 134° = 0.80328514559 rad
∠ C' = γ' = 106.6833350975° = 106°41' = 1.2879616922 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. Use the Law of Cosines

a=40 b=33 β=46  b2=a2+c22accosβ 332=402+c22 40 c cos(46)  c255.573c+511=0  p=1;q=55.573;r=511 D=q24pr=55.573241511=1044.32161055 D>0  c1,2=q±D2p=55.57±1044.322 c1,2=27.78633482±16.157982629 c1=43.9443174473 c2=11.6283521894   Factored form of the equation:  (c43.9443174473)(c11.6283521894)=0   c>0a = 40 \ \\ b = 33 \ \\ β = 46^\circ \ \\ \ \\ b^2 = a^2 + c^2 - 2ac \cos β \ \\ 33^2 = 40^2 + c^2 -2 \cdot \ 40 \cdot \ c \cdot \ \cos (46^\circ ) \ \\ \ \\ c^2 -55.573c +511 =0 \ \\ \ \\ p=1; q=-55.573; r=511 \ \\ D = q^2 - 4pr = 55.573^2 - 4\cdot 1 \cdot 511 = 1044.32161055 \ \\ D>0 \ \\ \ \\ c_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 55.57 \pm \sqrt{ 1044.32 } }{ 2 } \ \\ c_{1,2} = 27.78633482 \pm 16.157982629 \ \\ c_{1} = 43.9443174473 \ \\ c_{2} = 11.6283521894 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (c -43.9443174473) (c -11.6283521894) = 0 \ \\ \ \\ \ \\ c>0

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=40 b=33 c=43.94a = 40 \ \\ b = 33 \ \\ c = 43.94

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

p=a+b+c=40+33+43.94=116.94p = a+b+c = 40+33+43.94 = 116.94

3. 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=116.942=58.47s = \dfrac{ p }{ 2 } = \dfrac{ 116.94 }{ 2 } = 58.47

4. 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=58.47(58.4740)(58.4733)(58.4743.94) T=399699.51=632.22T = \sqrt{ s(s-a)(s-b)(s-c) } \ \\ T = \sqrt{ 58.47(58.47-40)(58.47-33)(58.47-43.94) } \ \\ T = \sqrt{ 399699.51 } = 632.22

5. 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=2 632.2240=31.61 hb=2 Tb=2 632.2233=38.32 hc=2 Tc=2 632.2243.94=28.77T = \dfrac{ a h _a }{ 2 } \ \\ \ \\ h _a = \dfrac{ 2 \ T }{ a } = \dfrac{ 2 \cdot \ 632.22 }{ 40 } = 31.61 \ \\ h _b = \dfrac{ 2 \ T }{ b } = \dfrac{ 2 \cdot \ 632.22 }{ 33 } = 38.32 \ \\ h _c = \dfrac{ 2 \ T }{ c } = \dfrac{ 2 \cdot \ 632.22 }{ 43.94 } = 28.77

6. 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(332+43.9424022 33 43.94)=6041  b2=a2+c22accosβ β=arccos(a2+c2b22ac)=arccos(402+43.9423322 40 43.94)=46 γ=180αβ=180604146=7319a^2 = b^2+c^2 - 2bc \cos α \ \\ \ \\ α = \arccos(\dfrac{ b^2+c^2-a^2 }{ 2bc } ) = \arccos(\dfrac{ 33^2+43.94^2-40^2 }{ 2 \cdot \ 33 \cdot \ 43.94 } ) = 60^\circ 41' \ \\ \ \\ b^2 = a^2+c^2 - 2ac \cos β \ \\ β = \arccos(\dfrac{ a^2+c^2-b^2 }{ 2ac } ) = \arccos(\dfrac{ 40^2+43.94^2-33^2 }{ 2 \cdot \ 40 \cdot \ 43.94 } ) = 46^\circ \ \\ γ = 180^\circ - α - β = 180^\circ - 60^\circ 41' - 46^\circ = 73^\circ 19'

7. 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=632.2258.47=10.81T = rs \ \\ r = \dfrac{ T }{ s } = \dfrac{ 632.22 }{ 58.47 } = 10.81

8. 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=40 33 43.944 10.812 58.472=22.94R = \dfrac{ a b c }{ 4 \ r s } = \dfrac{ 40 \cdot \ 33 \cdot \ 43.94 }{ 4 \cdot \ 10.812 \cdot \ 58.472 } = 22.94

9. 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=2 332+2 43.9424022=33.317 mb=2c2+2a2b22=2 43.942+2 4023322=38.643 mc=2a2+2b2c22=2 402+2 33243.9422=29.355m_a = \dfrac{ \sqrt{ 2b^2+2c^2 - a^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 33^2+2 \cdot \ 43.94^2 - 40^2 } }{ 2 } = 33.317 \ \\ m_b = \dfrac{ \sqrt{ 2c^2+2a^2 - b^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 43.94^2+2 \cdot \ 40^2 - 33^2 } }{ 2 } = 38.643 \ \\ m_c = \dfrac{ \sqrt{ 2a^2+2b^2 - c^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 40^2+2 \cdot \ 33^2 - 43.94^2 } }{ 2 } = 29.355



#2 Obtuse scalene triangle.

Sides: a = 40   b = 33   c = 11.62883521894

Area: T = 167.2954730844
Perimeter: p = 84.62883521894
Semiperimeter: s = 42.31441760947

Angle ∠ A = α = 119.3176649025° = 119°19' = 2.08224683779 rad
Angle ∠ B = β = 46° = 0.80328514559 rad
Angle ∠ C = γ = 14.68333509748° = 14°41' = 0.25662728197 rad

Height: ha = 8.36547365422
Height: hb = 10.13990745966
Height: hc = 28.77435920135

Median: ma = 14.5643972237
Median: mb = 24.43999853959
Median: mc = 36.20435268495

Inradius: r = 3.95436331859
Circumradius: R = 22.93876992518

Vertex coordinates: A[11.62883521894; 0] B[0; 0] C[27.78663348184; 28.77435920135]
Centroid: CG[13.13882290026; 9.59111973378]
Coordinates of the circumscribed circle: U[5.81441760947; 22.18985872309]
Coordinates of the inscribed circle: I[9.31441760947; 3.95436331859]

Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 60.68333509748° = 60°41' = 2.08224683779 rad
∠ B' = β' = 134° = 0.80328514559 rad
∠ C' = γ' = 165.3176649025° = 165°19' = 0.25662728197 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. Use the Law of Cosines

a=40 b=33 β=46  b2=a2+c22accosβ 332=402+c22 40 c cos(46)  c255.573c+511=0  p=1;q=55.573;r=511 D=q24pr=55.573241511=1044.32161055 D>0  c1,2=q±D2p=55.57±1044.322 c1,2=27.78633482±16.157982629 c1=43.9443174473 c2=11.6283521894   Factored form of the equation:  (c43.9443174473)(c11.6283521894)=0   c>0a = 40 \ \\ b = 33 \ \\ β = 46^\circ \ \\ \ \\ b^2 = a^2 + c^2 - 2ac \cos β \ \\ 33^2 = 40^2 + c^2 -2 \cdot \ 40 \cdot \ c \cdot \ \cos (46^\circ ) \ \\ \ \\ c^2 -55.573c +511 =0 \ \\ \ \\ p=1; q=-55.573; r=511 \ \\ D = q^2 - 4pr = 55.573^2 - 4\cdot 1 \cdot 511 = 1044.32161055 \ \\ D>0 \ \\ \ \\ c_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 55.57 \pm \sqrt{ 1044.32 } }{ 2 } \ \\ c_{1,2} = 27.78633482 \pm 16.157982629 \ \\ c_{1} = 43.9443174473 \ \\ c_{2} = 11.6283521894 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (c -43.9443174473) (c -11.6283521894) = 0 \ \\ \ \\ \ \\ c>0

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=40 b=33 c=11.63a = 40 \ \\ b = 33 \ \\ c = 11.63

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

p=a+b+c=40+33+11.63=84.63p = a+b+c = 40+33+11.63 = 84.63

3. 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=84.632=42.31s = \dfrac{ p }{ 2 } = \dfrac{ 84.63 }{ 2 } = 42.31

4. 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=42.31(42.3140)(42.3133)(42.3111.63) T=27987.53=167.29T = \sqrt{ s(s-a)(s-b)(s-c) } \ \\ T = \sqrt{ 42.31(42.31-40)(42.31-33)(42.31-11.63) } \ \\ T = \sqrt{ 27987.53 } = 167.29

5. 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=2 167.2940=8.36 hb=2 Tb=2 167.2933=10.14 hc=2 Tc=2 167.2911.63=28.77T = \dfrac{ a h _a }{ 2 } \ \\ \ \\ h _a = \dfrac{ 2 \ T }{ a } = \dfrac{ 2 \cdot \ 167.29 }{ 40 } = 8.36 \ \\ h _b = \dfrac{ 2 \ T }{ b } = \dfrac{ 2 \cdot \ 167.29 }{ 33 } = 10.14 \ \\ h _c = \dfrac{ 2 \ T }{ c } = \dfrac{ 2 \cdot \ 167.29 }{ 11.63 } = 28.77

6. 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(332+11.6324022 33 11.63)=11919  b2=a2+c22accosβ β=arccos(a2+c2b22ac)=arccos(402+11.6323322 40 11.63)=46 γ=180αβ=1801191946=1441a^2 = b^2+c^2 - 2bc \cos α \ \\ \ \\ α = \arccos(\dfrac{ b^2+c^2-a^2 }{ 2bc } ) = \arccos(\dfrac{ 33^2+11.63^2-40^2 }{ 2 \cdot \ 33 \cdot \ 11.63 } ) = 119^\circ 19' \ \\ \ \\ b^2 = a^2+c^2 - 2ac \cos β \ \\ β = \arccos(\dfrac{ a^2+c^2-b^2 }{ 2ac } ) = \arccos(\dfrac{ 40^2+11.63^2-33^2 }{ 2 \cdot \ 40 \cdot \ 11.63 } ) = 46^\circ \ \\ γ = 180^\circ - α - β = 180^\circ - 119^\circ 19' - 46^\circ = 14^\circ 41'

7. 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=167.2942.31=3.95T = rs \ \\ r = \dfrac{ T }{ s } = \dfrac{ 167.29 }{ 42.31 } = 3.95

8. 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=40 33 11.634 3.954 42.314=22.94R = \dfrac{ a b c }{ 4 \ r s } = \dfrac{ 40 \cdot \ 33 \cdot \ 11.63 }{ 4 \cdot \ 3.954 \cdot \ 42.314 } = 22.94

9. 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=2 332+2 11.6324022=14.564 mb=2c2+2a2b22=2 11.632+2 4023322=24.4 mc=2a2+2b2c22=2 402+2 33211.6322=36.204m_a = \dfrac{ \sqrt{ 2b^2+2c^2 - a^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 33^2+2 \cdot \ 11.63^2 - 40^2 } }{ 2 } = 14.564 \ \\ m_b = \dfrac{ \sqrt{ 2c^2+2a^2 - b^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 11.63^2+2 \cdot \ 40^2 - 33^2 } }{ 2 } = 24.4 \ \\ m_c = \dfrac{ \sqrt{ 2a^2+2b^2 - c^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 40^2+2 \cdot \ 33^2 - 11.63^2 } }{ 2 } = 36.204

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