Triangle calculator SSA

Please enter two sides and a non-included angle
°


Triangle has two solutions with side c=122.0411264747 and with side c=41.78991441111

#1 Acute scalene triangle.

Sides: a = 100   b = 70   c = 122.0411264747

Area: T = 35009.99968606
Perimeter: p = 292.0411264747
Semiperimeter: s = 146.0210632373

Angle ∠ A = α = 55.02442676241° = 55°1'27″ = 0.96603546385 rad
Angle ∠ B = β = 35° = 0.61108652382 rad
Angle ∠ C = γ = 89.97657323759° = 89°58'33″ = 1.57703727769 rad

Height: ha = 709.9999937212
Height: hb = 100.9999910303
Height: hc = 57.35876436351

Median: ma = 86.00660181062
Median: mb = 105.9344107588
Median: mc = 61.04549213674

Inradius: r = 23.96992133171
Circumradius: R = 61.02106378467

Vertex coordinates: A[122.0411264747; 0] B[0; 0] C[81.91552044289; 57.35876436351]
Centroid: CG[67.98554897252; 19.1199214545]
Coordinates of the circumscribed circle: U[61.02106323734; 0.02658452869]
Coordinates of the inscribed circle: I[76.02106323734; 23.96992133171]

Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 124.9765732376° = 124°58'33″ = 0.96603546385 rad
∠ B' = β' = 145° = 0.61108652382 rad
∠ C' = γ' = 90.02442676241° = 90°1'27″ = 1.57703727769 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=100 b=70 β=35  b2=a2+c22accosβ 702=1002+c22 100 c cos(35)  c2163.83c+5100=0  p=1;q=163.83;r=5100 D=q24pr=163.832415100=6440.40286651 D>0  c1,2=q±D2p=163.83±6440.42 c1,2=81.91520443±40.1260603178 c1=122.041264747 c2=41.7891441111   Factored form of the equation:  (c122.041264747)(c41.7891441111)=0   c>0a = 100 \ \\ b = 70 \ \\ β = 35^\circ \ \\ \ \\ b^2 = a^2 + c^2 - 2ac \cos β \ \\ 70^2 = 100^2 + c^2 -2 \cdot \ 100 \cdot \ c \cdot \ \cos (35^\circ ) \ \\ \ \\ c^2 -163.83c +5100 =0 \ \\ \ \\ p=1; q=-163.83; r=5100 \ \\ D = q^2 - 4pr = 163.83^2 - 4\cdot 1 \cdot 5100 = 6440.40286651 \ \\ D>0 \ \\ \ \\ c_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 163.83 \pm \sqrt{ 6440.4 } }{ 2 } \ \\ c_{1,2} = 81.91520443 \pm 40.1260603178 \ \\ c_{1} = 122.041264747 \ \\ c_{2} = 41.7891441111 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (c -122.041264747) (c -41.7891441111) = 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=100 b=70 c=122.04a = 100 \ \\ b = 70 \ \\ c = 122.04

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

p=a+b+c=100+70+122.04=292.04p = a+b+c = 100+70+122.04 = 292.04

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=292.042=146.02s = \dfrac{ p }{ 2 } = \dfrac{ 292.04 }{ 2 } = 146.02

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=146.02(146.02100)(146.0270)(146.02122.04) T=12249997.8=3500T = \sqrt{ s(s-a)(s-b)(s-c) } \ \\ T = \sqrt{ 146.02(146.02-100)(146.02-70)(146.02-122.04) } \ \\ T = \sqrt{ 12249997.8 } = 3500

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 3500100=70 hb=2 Tb=2 350070=100 hc=2 Tc=2 3500122.04=57.36T = \dfrac{ a h _a }{ 2 } \ \\ \ \\ h _a = \dfrac{ 2 \ T }{ a } = \dfrac{ 2 \cdot \ 3500 }{ 100 } = 70 \ \\ h _b = \dfrac{ 2 \ T }{ b } = \dfrac{ 2 \cdot \ 3500 }{ 70 } = 100 \ \\ h _c = \dfrac{ 2 \ T }{ c } = \dfrac{ 2 \cdot \ 3500 }{ 122.04 } = 57.36

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(702+122.04210022 70 122.04)=55127"  b2=a2+c22accosβ β=arccos(a2+c2b22ac)=arccos(1002+122.0427022 100 122.04)=35 γ=180αβ=18055127"35=895833"a^2 = b^2+c^2 - 2bc \cos α \ \\ \ \\ α = \arccos(\dfrac{ b^2+c^2-a^2 }{ 2bc } ) = \arccos(\dfrac{ 70^2+122.04^2-100^2 }{ 2 \cdot \ 70 \cdot \ 122.04 } ) = 55^\circ 1'27" \ \\ \ \\ b^2 = a^2+c^2 - 2ac \cos β \ \\ β = \arccos(\dfrac{ a^2+c^2-b^2 }{ 2ac } ) = \arccos(\dfrac{ 100^2+122.04^2-70^2 }{ 2 \cdot \ 100 \cdot \ 122.04 } ) = 35^\circ \ \\ γ = 180^\circ - α - β = 180^\circ - 55^\circ 1'27" - 35^\circ = 89^\circ 58'33"

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=3500146.02=23.97T = rs \ \\ r = \dfrac{ T }{ s } = \dfrac{ 3500 }{ 146.02 } = 23.97

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=100 70 122.044 23.969 146.021=61.02R = \dfrac{ a b c }{ 4 \ r s } = \dfrac{ 100 \cdot \ 70 \cdot \ 122.04 }{ 4 \cdot \ 23.969 \cdot \ 146.021 } = 61.02

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=2 702+2 122.04210022=86.006 mb=2c2+2a2b22=2 122.042+2 10027022=105.934 mc=2a2+2b2c22=2 1002+2 702122.0422=61.045m_a = \dfrac{ \sqrt{ 2b^2+2c^2 - a^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 70^2+2 \cdot \ 122.04^2 - 100^2 } }{ 2 } = 86.006 \ \\ m_b = \dfrac{ \sqrt{ 2c^2+2a^2 - b^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 122.04^2+2 \cdot \ 100^2 - 70^2 } }{ 2 } = 105.934 \ \\ m_c = \dfrac{ \sqrt{ 2a^2+2b^2 - c^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 100^2+2 \cdot \ 70^2 - 122.04^2 } }{ 2 } = 61.045



#2 Obtuse scalene triangle.

Sides: a = 100   b = 70   c = 41.78991441111

Area: T = 1198.463341787
Perimeter: p = 211.7899144111
Semiperimeter: s = 105.8954572056

Angle ∠ A = α = 124.9765732376° = 124°58'33″ = 2.1811238015 rad
Angle ∠ B = β = 35° = 0.61108652382 rad
Angle ∠ C = γ = 20.02442676241° = 20°1'27″ = 0.34994894003 rad

Height: ha = 23.96992683574
Height: hb = 34.24218119391
Height: hc = 57.35876436351

Median: ma = 28.69108745557
Median: mb = 68.1777461692
Median: mc = 83.74661453359

Inradius: r = 11.31875150964
Circumradius: R = 61.02106378467

Vertex coordinates: A[41.78991441111; 0] B[0; 0] C[81.91552044289; 57.35876436351]
Centroid: CG[41.23547828467; 19.1199214545]
Coordinates of the circumscribed circle: U[20.89545720555; 57.33217983482]
Coordinates of the inscribed circle: I[35.89545720555; 11.31875150964]

Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 55.02442676241° = 55°1'27″ = 2.1811238015 rad
∠ B' = β' = 145° = 0.61108652382 rad
∠ C' = γ' = 159.9765732376° = 159°58'33″ = 0.34994894003 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=100 b=70 β=35  b2=a2+c22accosβ 702=1002+c22 100 c cos(35)  c2163.83c+5100=0  p=1;q=163.83;r=5100 D=q24pr=163.832415100=6440.40286651 D>0  c1,2=q±D2p=163.83±6440.42 c1,2=81.91520443±40.1260603178 c1=122.041264747 c2=41.7891441111   Factored form of the equation:  (c122.041264747)(c41.7891441111)=0   c>0a = 100 \ \\ b = 70 \ \\ β = 35^\circ \ \\ \ \\ b^2 = a^2 + c^2 - 2ac \cos β \ \\ 70^2 = 100^2 + c^2 -2 \cdot \ 100 \cdot \ c \cdot \ \cos (35^\circ ) \ \\ \ \\ c^2 -163.83c +5100 =0 \ \\ \ \\ p=1; q=-163.83; r=5100 \ \\ D = q^2 - 4pr = 163.83^2 - 4\cdot 1 \cdot 5100 = 6440.40286651 \ \\ D>0 \ \\ \ \\ c_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 163.83 \pm \sqrt{ 6440.4 } }{ 2 } \ \\ c_{1,2} = 81.91520443 \pm 40.1260603178 \ \\ c_{1} = 122.041264747 \ \\ c_{2} = 41.7891441111 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (c -122.041264747) (c -41.7891441111) = 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=100 b=70 c=41.79a = 100 \ \\ b = 70 \ \\ c = 41.79

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

p=a+b+c=100+70+41.79=211.79p = a+b+c = 100+70+41.79 = 211.79

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=211.792=105.89s = \dfrac{ p }{ 2 } = \dfrac{ 211.79 }{ 2 } = 105.89

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=105.89(105.89100)(105.8970)(105.8941.79) T=1436314.56=1198.46T = \sqrt{ s(s-a)(s-b)(s-c) } \ \\ T = \sqrt{ 105.89(105.89-100)(105.89-70)(105.89-41.79) } \ \\ T = \sqrt{ 1436314.56 } = 1198.46

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 1198.46100=23.97 hb=2 Tb=2 1198.4670=34.24 hc=2 Tc=2 1198.4641.79=57.36T = \dfrac{ a h _a }{ 2 } \ \\ \ \\ h _a = \dfrac{ 2 \ T }{ a } = \dfrac{ 2 \cdot \ 1198.46 }{ 100 } = 23.97 \ \\ h _b = \dfrac{ 2 \ T }{ b } = \dfrac{ 2 \cdot \ 1198.46 }{ 70 } = 34.24 \ \\ h _c = \dfrac{ 2 \ T }{ c } = \dfrac{ 2 \cdot \ 1198.46 }{ 41.79 } = 57.36

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(702+41.79210022 70 41.79)=1245833"  b2=a2+c22accosβ β=arccos(a2+c2b22ac)=arccos(1002+41.7927022 100 41.79)=35 γ=180αβ=1801245833"35=20127"a^2 = b^2+c^2 - 2bc \cos α \ \\ \ \\ α = \arccos(\dfrac{ b^2+c^2-a^2 }{ 2bc } ) = \arccos(\dfrac{ 70^2+41.79^2-100^2 }{ 2 \cdot \ 70 \cdot \ 41.79 } ) = 124^\circ 58'33" \ \\ \ \\ b^2 = a^2+c^2 - 2ac \cos β \ \\ β = \arccos(\dfrac{ a^2+c^2-b^2 }{ 2ac } ) = \arccos(\dfrac{ 100^2+41.79^2-70^2 }{ 2 \cdot \ 100 \cdot \ 41.79 } ) = 35^\circ \ \\ γ = 180^\circ - α - β = 180^\circ - 124^\circ 58'33" - 35^\circ = 20^\circ 1'27"

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=1198.46105.89=11.32T = rs \ \\ r = \dfrac{ T }{ s } = \dfrac{ 1198.46 }{ 105.89 } = 11.32

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=100 70 41.794 11.318 105.895=61.02R = \dfrac{ a b c }{ 4 \ r s } = \dfrac{ 100 \cdot \ 70 \cdot \ 41.79 }{ 4 \cdot \ 11.318 \cdot \ 105.895 } = 61.02

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=2 702+2 41.79210022=28.691 mb=2c2+2a2b22=2 41.792+2 10027022=68.177 mc=2a2+2b2c22=2 1002+2 70241.7922=83.746m_a = \dfrac{ \sqrt{ 2b^2+2c^2 - a^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 70^2+2 \cdot \ 41.79^2 - 100^2 } }{ 2 } = 28.691 \ \\ m_b = \dfrac{ \sqrt{ 2c^2+2a^2 - b^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 41.79^2+2 \cdot \ 100^2 - 70^2 } }{ 2 } = 68.177 \ \\ m_c = \dfrac{ \sqrt{ 2a^2+2b^2 - c^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 100^2+2 \cdot \ 70^2 - 41.79^2 } }{ 2 } = 83.746

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