Triangle calculator SSA

Please enter two sides and a non-included angle
°


Triangle has two solutions with side c=74.49548974278 and with side c=25.50551025722

#1 Acute scalene triangle.

Sides: a = 100   b = 90   c = 74.49548974278

Area: T = 3225.724368124
Perimeter: p = 264.4954897428
Semiperimeter: s = 132.2477448714

Angle ∠ A = α = 74.20768309517° = 74°12'25″ = 1.29551535276 rad
Angle ∠ B = β = 60° = 1.04771975512 rad
Angle ∠ C = γ = 45.79331690483° = 45°47'35″ = 0.79992415748 rad

Height: ha = 64.51444736248
Height: hb = 71.6832748472
Height: hc = 86.60325403784

Median: ma = 65.76327924543
Median: mb = 75.82770721536
Median: mc = 87.53664356386

Inradius: r = 24.39215758876
Circumradius: R = 51.96215242271

Vertex coordinates: A[74.49548974278; 0] B[0; 0] C[50; 86.60325403784]
Centroid: CG[41.49882991426; 28.86875134595]
Coordinates of the circumscribed circle: U[37.24774487139; 36.23302023774]
Coordinates of the inscribed circle: I[42.24774487139; 24.39215758876]

Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 105.7933169048° = 105°47'35″ = 1.29551535276 rad
∠ B' = β' = 120° = 1.04771975512 rad
∠ C' = γ' = 134.2076830952° = 134°12'25″ = 0.79992415748 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=90 β=60  b2=a2+c22accosβ 902=1002+c22 100 c cos(60)  c2100c+1900=0  p=1;q=100;r=1900 D=q24pr=1002411900=2400 D>0  c1,2=q±D2p=100±24002=50±600 c1,2=50±24.4948974278 c1=74.4948974278 c2=25.5051025722   Factored form of the equation:  (c74.4948974278)(c25.5051025722)=0   c>0a = 100 \ \\ b = 90 \ \\ β = 60^\circ \ \\ \ \\ b^2 = a^2 + c^2 - 2ac \cos β \ \\ 90^2 = 100^2 + c^2 -2 \cdot \ 100 \cdot \ c \cdot \ \cos (60^\circ ) \ \\ \ \\ c^2 -100c +1900 =0 \ \\ \ \\ p=1; q=-100; r=1900 \ \\ D = q^2 - 4pr = 100^2 - 4\cdot 1 \cdot 1900 = 2400 \ \\ D>0 \ \\ \ \\ c_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 100 \pm \sqrt{ 2400 } }{ 2 } = 50 \pm \sqrt{ 600 } \ \\ c_{1,2} = 50 \pm 24.4948974278 \ \\ c_{1} = 74.4948974278 \ \\ c_{2} = 25.5051025722 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (c -74.4948974278) (c -25.5051025722) = 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=90 c=74.49a = 100 \ \\ b = 90 \ \\ c = 74.49

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

p=a+b+c=100+90+74.49=264.49p = a+b+c = 100+90+74.49 = 264.49

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=264.492=132.25s = \dfrac{ p }{ 2 } = \dfrac{ 264.49 }{ 2 } = 132.25

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=132.25(132.25100)(132.2590)(132.2574.49) T=10405293.27=3225.72T = \sqrt{ s(s-a)(s-b)(s-c) } \ \\ T = \sqrt{ 132.25(132.25-100)(132.25-90)(132.25-74.49) } \ \\ T = \sqrt{ 10405293.27 } = 3225.72

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 3225.72100=64.51 hb=2 Tb=2 3225.7290=71.68 hc=2 Tc=2 3225.7274.49=86.6T = \dfrac{ a h _a }{ 2 } \ \\ \ \\ h _a = \dfrac{ 2 \ T }{ a } = \dfrac{ 2 \cdot \ 3225.72 }{ 100 } = 64.51 \ \\ h _b = \dfrac{ 2 \ T }{ b } = \dfrac{ 2 \cdot \ 3225.72 }{ 90 } = 71.68 \ \\ h _c = \dfrac{ 2 \ T }{ c } = \dfrac{ 2 \cdot \ 3225.72 }{ 74.49 } = 86.6

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+74.49210022 90 74.49)=741225"  b2=a2+c22accosβ β=arccos(a2+c2b22ac)=arccos(1002+74.4929022 100 74.49)=60 γ=180αβ=180741225"60=454735"a^2 = b^2+c^2 - 2bc \cos α \ \\ \ \\ α = \arccos(\dfrac{ b^2+c^2-a^2 }{ 2bc } ) = \arccos(\dfrac{ 90^2+74.49^2-100^2 }{ 2 \cdot \ 90 \cdot \ 74.49 } ) = 74^\circ 12'25" \ \\ \ \\ b^2 = a^2+c^2 - 2ac \cos β \ \\ β = \arccos(\dfrac{ a^2+c^2-b^2 }{ 2ac } ) = \arccos(\dfrac{ 100^2+74.49^2-90^2 }{ 2 \cdot \ 100 \cdot \ 74.49 } ) = 60^\circ \ \\ γ = 180^\circ - α - β = 180^\circ - 74^\circ 12'25" - 60^\circ = 45^\circ 47'35"

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=3225.72132.25=24.39T = rs \ \\ r = \dfrac{ T }{ s } = \dfrac{ 3225.72 }{ 132.25 } = 24.39

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 74.494 24.392 132.247=51.96R = \dfrac{ a b c }{ 4 \ r s } = \dfrac{ 100 \cdot \ 90 \cdot \ 74.49 }{ 4 \cdot \ 24.392 \cdot \ 132.247 } = 51.96

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 74.49210022=65.763 mb=2c2+2a2b22=2 74.492+2 10029022=75.827 mc=2a2+2b2c22=2 1002+2 90274.4922=87.536m_a = \dfrac{ \sqrt{ 2b^2+2c^2 - a^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 90^2+2 \cdot \ 74.49^2 - 100^2 } }{ 2 } = 65.763 \ \\ m_b = \dfrac{ \sqrt{ 2c^2+2a^2 - b^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 74.49^2+2 \cdot \ 100^2 - 90^2 } }{ 2 } = 75.827 \ \\ m_c = \dfrac{ \sqrt{ 2a^2+2b^2 - c^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 100^2+2 \cdot \ 90^2 - 74.49^2 } }{ 2 } = 87.536



#2 Obtuse scalene triangle.

Sides: a = 100   b = 90   c = 25.50551025722

Area: T = 1104.403333768
Perimeter: p = 215.5055102572
Semiperimeter: s = 107.7532551286

Angle ∠ A = α = 105.7933169048° = 105°47'35″ = 1.8466439126 rad
Angle ∠ B = β = 60° = 1.04771975512 rad
Angle ∠ C = γ = 14.20768309517° = 14°12'25″ = 0.24879559764 rad

Height: ha = 22.08880667536
Height: hb = 24.54222963929
Height: hc = 86.60325403784

Median: ma = 43.30442160604
Median: mb = 57.4487847032
Median: mc = 94.27328616077

Inradius: r = 10.24994402638
Circumradius: R = 51.96215242271

Vertex coordinates: A[25.50551025722; 0] B[0; 0] C[50; 86.60325403784]
Centroid: CG[25.16883675241; 28.86875134595]
Coordinates of the circumscribed circle: U[12.75325512861; 50.37223380011]
Coordinates of the inscribed circle: I[17.75325512861; 10.24994402638]

Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 74.20768309517° = 74°12'25″ = 1.8466439126 rad
∠ B' = β' = 120° = 1.04771975512 rad
∠ C' = γ' = 165.7933169048° = 165°47'35″ = 0.24879559764 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=90 β=60  b2=a2+c22accosβ 902=1002+c22 100 c cos(60)  c2100c+1900=0  p=1;q=100;r=1900 D=q24pr=1002411900=2400 D>0  c1,2=q±D2p=100±24002=50±600 c1,2=50±24.4948974278 c1=74.4948974278 c2=25.5051025722   Factored form of the equation:  (c74.4948974278)(c25.5051025722)=0   c>0a = 100 \ \\ b = 90 \ \\ β = 60^\circ \ \\ \ \\ b^2 = a^2 + c^2 - 2ac \cos β \ \\ 90^2 = 100^2 + c^2 -2 \cdot \ 100 \cdot \ c \cdot \ \cos (60^\circ ) \ \\ \ \\ c^2 -100c +1900 =0 \ \\ \ \\ p=1; q=-100; r=1900 \ \\ D = q^2 - 4pr = 100^2 - 4\cdot 1 \cdot 1900 = 2400 \ \\ D>0 \ \\ \ \\ c_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 100 \pm \sqrt{ 2400 } }{ 2 } = 50 \pm \sqrt{ 600 } \ \\ c_{1,2} = 50 \pm 24.4948974278 \ \\ c_{1} = 74.4948974278 \ \\ c_{2} = 25.5051025722 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (c -74.4948974278) (c -25.5051025722) = 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=90 c=25.51a = 100 \ \\ b = 90 \ \\ c = 25.51

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

p=a+b+c=100+90+25.51=215.51p = a+b+c = 100+90+25.51 = 215.51

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=215.512=107.75s = \dfrac{ p }{ 2 } = \dfrac{ 215.51 }{ 2 } = 107.75

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=107.75(107.75100)(107.7590)(107.7525.51) T=1219706.73=1104.4T = \sqrt{ s(s-a)(s-b)(s-c) } \ \\ T = \sqrt{ 107.75(107.75-100)(107.75-90)(107.75-25.51) } \ \\ T = \sqrt{ 1219706.73 } = 1104.4

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 1104.4100=22.09 hb=2 Tb=2 1104.490=24.54 hc=2 Tc=2 1104.425.51=86.6T = \dfrac{ a h _a }{ 2 } \ \\ \ \\ h _a = \dfrac{ 2 \ T }{ a } = \dfrac{ 2 \cdot \ 1104.4 }{ 100 } = 22.09 \ \\ h _b = \dfrac{ 2 \ T }{ b } = \dfrac{ 2 \cdot \ 1104.4 }{ 90 } = 24.54 \ \\ h _c = \dfrac{ 2 \ T }{ c } = \dfrac{ 2 \cdot \ 1104.4 }{ 25.51 } = 86.6

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+25.51210022 90 25.51)=1054735"  b2=a2+c22accosβ β=arccos(a2+c2b22ac)=arccos(1002+25.5129022 100 25.51)=60 γ=180αβ=1801054735"60=141225"a^2 = b^2+c^2 - 2bc \cos α \ \\ \ \\ α = \arccos(\dfrac{ b^2+c^2-a^2 }{ 2bc } ) = \arccos(\dfrac{ 90^2+25.51^2-100^2 }{ 2 \cdot \ 90 \cdot \ 25.51 } ) = 105^\circ 47'35" \ \\ \ \\ b^2 = a^2+c^2 - 2ac \cos β \ \\ β = \arccos(\dfrac{ a^2+c^2-b^2 }{ 2ac } ) = \arccos(\dfrac{ 100^2+25.51^2-90^2 }{ 2 \cdot \ 100 \cdot \ 25.51 } ) = 60^\circ \ \\ γ = 180^\circ - α - β = 180^\circ - 105^\circ 47'35" - 60^\circ = 14^\circ 12'25"

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=1104.4107.75=10.25T = rs \ \\ r = \dfrac{ T }{ s } = \dfrac{ 1104.4 }{ 107.75 } = 10.25

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 25.514 10.249 107.753=51.96R = \dfrac{ a b c }{ 4 \ r s } = \dfrac{ 100 \cdot \ 90 \cdot \ 25.51 }{ 4 \cdot \ 10.249 \cdot \ 107.753 } = 51.96

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 25.51210022=43.304 mb=2c2+2a2b22=2 25.512+2 10029022=57.448 mc=2a2+2b2c22=2 1002+2 90225.5122=94.273m_a = \dfrac{ \sqrt{ 2b^2+2c^2 - a^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 90^2+2 \cdot \ 25.51^2 - 100^2 } }{ 2 } = 43.304 \ \\ m_b = \dfrac{ \sqrt{ 2c^2+2a^2 - b^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 25.51^2+2 \cdot \ 100^2 - 90^2 } }{ 2 } = 57.448 \ \\ m_c = \dfrac{ \sqrt{ 2a^2+2b^2 - c^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 100^2+2 \cdot \ 90^2 - 25.51^2 } }{ 2 } = 94.273

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