Triangle calculator SSA

Please enter two sides and a non-included angle
°


Triangle has two solutions with side c=161.4365688114 and with side c=11.7699392643

#1 Obtuse scalene triangle.

Sides: a = 100   b = 90   c = 161.4365688114

Area: T = 4035.892220285
Perimeter: p = 351.4365688114
Semiperimeter: s = 175.7187844057

Angle ∠ A = α = 33.74989885959° = 33°44'56″ = 0.58990309702 rad
Angle ∠ B = β = 30° = 0.52435987756 rad
Angle ∠ C = γ = 116.2511011404° = 116°15'4″ = 2.02989629078 rad

Height: ha = 80.7187844057
Height: hb = 89.68664933966
Height: hc = 50

Median: ma = 120.7510737879
Median: mb = 126.5143796475
Median: mc = 50.34551055297

Inradius: r = 22.9688027092
Circumradius: R = 90

Vertex coordinates: A[161.4365688114; 0] B[0; 0] C[86.60325403784; 50]
Centroid: CG[82.67994094975; 16.66766666667]
Coordinates of the circumscribed circle: U[80.7187844057; -39.80774069841]
Coordinates of the inscribed circle: I[85.7187844057; 22.9688027092]

Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 146.2511011404° = 146°15'4″ = 0.58990309702 rad
∠ B' = β' = 150° = 0.52435987756 rad
∠ C' = γ' = 63.74989885959° = 63°44'56″ = 2.02989629078 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


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=90 c=161.44a = 100 \ \\ b = 90 \ \\ c = 161.44

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

p=a+b+c=100+90+161.44=351.44p = a+b+c = 100+90+161.44 = 351.44

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=351.442=175.72s = \dfrac{ p }{ 2 } = \dfrac{ 351.44 }{ 2 } = 175.72

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=175.72(175.72100)(175.7290)(175.72161.44) T=16288425.87=4035.89T = \sqrt{ s(s-a)(s-b)(s-c) } \ \\ T = \sqrt{ 175.72(175.72-100)(175.72-90)(175.72-161.44) } \ \\ T = \sqrt{ 16288425.87 } = 4035.89

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 4035.89100=80.72 hb=2 Tb=2 4035.8990=89.69 hc=2 Tc=2 4035.89161.44=50T = \dfrac{ a h _a }{ 2 } \ \\ \ \\ h _a = \dfrac{ 2 \ T }{ a } = \dfrac{ 2 \cdot \ 4035.89 }{ 100 } = 80.72 \ \\ h _b = \dfrac{ 2 \ T }{ b } = \dfrac{ 2 \cdot \ 4035.89 }{ 90 } = 89.69 \ \\ h _c = \dfrac{ 2 \ T }{ c } = \dfrac{ 2 \cdot \ 4035.89 }{ 161.44 } = 50

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(902+161.44210022 90 161.44)=334456"  b2=a2+c22accosβ β=arccos(a2+c2b22ac)=arccos(1002+161.4429022 100 161.44)=30 γ=180αβ=180334456"30=116154"a^2 = b^2+c^2 - 2bc \cos α \ \\ \ \\ α = \arccos(\dfrac{ b^2+c^2-a^2 }{ 2bc } ) = \arccos(\dfrac{ 90^2+161.44^2-100^2 }{ 2 \cdot \ 90 \cdot \ 161.44 } ) = 33^\circ 44'56" \ \\ \ \\ b^2 = a^2+c^2 - 2ac \cos β \ \\ β = \arccos(\dfrac{ a^2+c^2-b^2 }{ 2ac } ) = \arccos(\dfrac{ 100^2+161.44^2-90^2 }{ 2 \cdot \ 100 \cdot \ 161.44 } ) = 30^\circ \ \\ γ = 180^\circ - α - β = 180^\circ - 33^\circ 44'56" - 30^\circ = 116^\circ 15'4"

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=4035.89175.72=22.97T = rs \ \\ r = \dfrac{ T }{ s } = \dfrac{ 4035.89 }{ 175.72 } = 22.97

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 90 161.444 22.968 175.718=90R = \dfrac{ a b c }{ 4 \ r s } = \dfrac{ 100 \cdot \ 90 \cdot \ 161.44 }{ 4 \cdot \ 22.968 \cdot \ 175.718 } = 90

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=2 902+2 161.44210022=120.751 mb=2c2+2a2b22=2 161.442+2 10029022=126.514 mc=2a2+2b2c22=2 1002+2 902161.4422=50.345m_a = \dfrac{ \sqrt{ 2b^2+2c^2 - a^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 90^2+2 \cdot \ 161.44^2 - 100^2 } }{ 2 } = 120.751 \ \\ m_b = \dfrac{ \sqrt{ 2c^2+2a^2 - b^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 161.44^2+2 \cdot \ 100^2 - 90^2 } }{ 2 } = 126.514 \ \\ m_c = \dfrac{ \sqrt{ 2a^2+2b^2 - c^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 100^2+2 \cdot \ 90^2 - 161.44^2 } }{ 2 } = 50.345



#2 Obtuse scalene triangle.

Sides: a = 100   b = 90   c = 11.7699392643

Area: T = 294.2354816074
Perimeter: p = 201.7699392643
Semiperimeter: s = 100.8854696322

Angle ∠ A = α = 146.2511011404° = 146°15'4″ = 2.55325616834 rad
Angle ∠ B = β = 30° = 0.52435987756 rad
Angle ∠ C = γ = 3.74989885959° = 3°44'56″ = 0.06554321946 rad

Height: ha = 5.88546963215
Height: hb = 6.53985514683
Height: hc = 50

Median: ma = 40.2440021143
Median: mb = 55.17548067653
Median: mc = 94.94993041007

Inradius: r = 2.91765455892
Circumradius: R = 90

Vertex coordinates: A[11.7699392643; 0] B[0; 0] C[86.60325403784; 50]
Centroid: CG[32.79106443405; 16.66766666667]
Coordinates of the circumscribed circle: U[5.88546963215; 89.80774069841]
Coordinates of the inscribed circle: I[10.88546963215; 2.91765455892]

Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 33.74989885959° = 33°44'56″ = 2.55325616834 rad
∠ B' = β' = 150° = 0.52435987756 rad
∠ C' = γ' = 176.2511011404° = 176°15'4″ = 0.06554321946 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


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=90 c=11.77a = 100 \ \\ b = 90 \ \\ c = 11.77

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

p=a+b+c=100+90+11.77=201.77p = a+b+c = 100+90+11.77 = 201.77

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=201.772=100.88s = \dfrac{ p }{ 2 } = \dfrac{ 201.77 }{ 2 } = 100.88

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=100.88(100.88100)(100.8890)(100.8811.77) T=86574.13=294.23T = \sqrt{ s(s-a)(s-b)(s-c) } \ \\ T = \sqrt{ 100.88(100.88-100)(100.88-90)(100.88-11.77) } \ \\ T = \sqrt{ 86574.13 } = 294.23

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=2 294.23100=5.88 hb=2 Tb=2 294.2390=6.54 hc=2 Tc=2 294.2311.77=50T = \dfrac{ a h _a }{ 2 } \ \\ \ \\ h _a = \dfrac{ 2 \ T }{ a } = \dfrac{ 2 \cdot \ 294.23 }{ 100 } = 5.88 \ \\ h _b = \dfrac{ 2 \ T }{ b } = \dfrac{ 2 \cdot \ 294.23 }{ 90 } = 6.54 \ \\ h _c = \dfrac{ 2 \ T }{ c } = \dfrac{ 2 \cdot \ 294.23 }{ 11.77 } = 50

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(902+11.77210022 90 11.77)=146154"  b2=a2+c22accosβ β=arccos(a2+c2b22ac)=arccos(1002+11.7729022 100 11.77)=30 γ=180αβ=180146154"30=34456"a^2 = b^2+c^2 - 2bc \cos α \ \\ \ \\ α = \arccos(\dfrac{ b^2+c^2-a^2 }{ 2bc } ) = \arccos(\dfrac{ 90^2+11.77^2-100^2 }{ 2 \cdot \ 90 \cdot \ 11.77 } ) = 146^\circ 15'4" \ \\ \ \\ b^2 = a^2+c^2 - 2ac \cos β \ \\ β = \arccos(\dfrac{ a^2+c^2-b^2 }{ 2ac } ) = \arccos(\dfrac{ 100^2+11.77^2-90^2 }{ 2 \cdot \ 100 \cdot \ 11.77 } ) = 30^\circ \ \\ γ = 180^\circ - α - β = 180^\circ - 146^\circ 15'4" - 30^\circ = 3^\circ 44'56"

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=294.23100.88=2.92T = rs \ \\ r = \dfrac{ T }{ s } = \dfrac{ 294.23 }{ 100.88 } = 2.92

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 90 11.774 2.917 100.885=90R = \dfrac{ a b c }{ 4 \ r s } = \dfrac{ 100 \cdot \ 90 \cdot \ 11.77 }{ 4 \cdot \ 2.917 \cdot \ 100.885 } = 90

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=2 902+2 11.77210022=40.24 mb=2c2+2a2b22=2 11.772+2 10029022=55.175 mc=2a2+2b2c22=2 1002+2 90211.7722=94.949m_a = \dfrac{ \sqrt{ 2b^2+2c^2 - a^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 90^2+2 \cdot \ 11.77^2 - 100^2 } }{ 2 } = 40.24 \ \\ m_b = \dfrac{ \sqrt{ 2c^2+2a^2 - b^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 11.77^2+2 \cdot \ 100^2 - 90^2 } }{ 2 } = 55.175 \ \\ m_c = \dfrac{ \sqrt{ 2a^2+2b^2 - c^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 100^2+2 \cdot \ 90^2 - 11.77^2 } }{ 2 } = 94.949

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