Triangle calculator SSA

Please enter two sides and a non-included angle
°


Triangle has two solutions with side c=48.76438049792 and with side c=5.90660198465

#1 Acute scalene triangle.

Sides: a = 38   b = 34   c = 48.76438049792

Area: T = 643.610952078
Perimeter: p = 120.7643804979
Semiperimeter: s = 60.38219024896

Angle ∠ A = α = 50.93105784325° = 50°55'50″ = 0.88989062836 rad
Angle ∠ B = β = 44° = 0.76879448709 rad
Angle ∠ C = γ = 85.06994215675° = 85°4'10″ = 1.48547414991 rad

Height: ha = 33.87441853042
Height: hb = 37.85993835753
Height: hc = 26.39770180774

Median: ma = 37.49660576331
Median: mb = 40.27334942366
Median: mc = 26.56216797471

Inradius: r = 10.65989804932
Circumradius: R = 24.47224611736

Vertex coordinates: A[48.76438049792; 0] B[0; 0] C[27.33549124129; 26.39770180774]
Centroid: CG[25.36662391307; 8.79990060258]
Coordinates of the circumscribed circle: U[24.38219024896; 2.1033375117]
Coordinates of the inscribed circle: I[26.38219024896; 10.65989804932]

Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 129.0699421568° = 129°4'10″ = 0.88989062836 rad
∠ B' = β' = 136° = 0.76879448709 rad
∠ C' = γ' = 94.93105784325° = 94°55'50″ = 1.48547414991 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=38 b=34 β=44  b2=a2+c22accosβ 342=382+c22 38 c cos(44)  c254.67c+288=0  p=1;q=54.67;r=288 D=q24pr=54.67241288=1836.78974648 D>0  c1,2=q±D2p=54.67±1836.792 c1,2=27.33491241±21.4288925663 c1=48.7638049792 c2=5.90601984654   Factored form of the equation:  (c48.7638049792)(c5.90601984654)=0   c>0a = 38 \ \\ b = 34 \ \\ β = 44^\circ \ \\ \ \\ b^2 = a^2 + c^2 - 2ac \cos β \ \\ 34^2 = 38^2 + c^2 -2 \cdot \ 38 \cdot \ c \cdot \ \cos (44^\circ ) \ \\ \ \\ c^2 -54.67c +288 =0 \ \\ \ \\ p=1; q=-54.67; r=288 \ \\ D = q^2 - 4pr = 54.67^2 - 4\cdot 1 \cdot 288 = 1836.78974648 \ \\ D>0 \ \\ \ \\ c_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 54.67 \pm \sqrt{ 1836.79 } }{ 2 } \ \\ c_{1,2} = 27.33491241 \pm 21.4288925663 \ \\ c_{1} = 48.7638049792 \ \\ c_{2} = 5.90601984654 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (c -48.7638049792) (c -5.90601984654) = 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=38 b=34 c=48.76a = 38 \ \\ b = 34 \ \\ c = 48.76

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

p=a+b+c=38+34+48.76=120.76p = a+b+c = 38+34+48.76 = 120.76

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=120.762=60.38s = \dfrac{ p }{ 2 } = \dfrac{ 120.76 }{ 2 } = 60.38

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=60.38(60.3838)(60.3834)(60.3848.76) T=414233.22=643.61T = \sqrt{ s(s-a)(s-b)(s-c) } \ \\ T = \sqrt{ 60.38(60.38-38)(60.38-34)(60.38-48.76) } \ \\ T = \sqrt{ 414233.22 } = 643.61

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=2 643.6138=33.87 hb=2 Tb=2 643.6134=37.86 hc=2 Tc=2 643.6148.76=26.4T = \dfrac{ a h _a }{ 2 } \ \\ \ \\ h _a = \dfrac{ 2 \ T }{ a } = \dfrac{ 2 \cdot \ 643.61 }{ 38 } = 33.87 \ \\ h _b = \dfrac{ 2 \ T }{ b } = \dfrac{ 2 \cdot \ 643.61 }{ 34 } = 37.86 \ \\ h _c = \dfrac{ 2 \ T }{ c } = \dfrac{ 2 \cdot \ 643.61 }{ 48.76 } = 26.4

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(342+48.7623822 34 48.76)=505550"  b2=a2+c22accosβ β=arccos(a2+c2b22ac)=arccos(382+48.7623422 38 48.76)=44 γ=180αβ=180505550"44=85410"a^2 = b^2+c^2 - 2bc \cos α \ \\ \ \\ α = \arccos(\dfrac{ b^2+c^2-a^2 }{ 2bc } ) = \arccos(\dfrac{ 34^2+48.76^2-38^2 }{ 2 \cdot \ 34 \cdot \ 48.76 } ) = 50^\circ 55'50" \ \\ \ \\ b^2 = a^2+c^2 - 2ac \cos β \ \\ β = \arccos(\dfrac{ a^2+c^2-b^2 }{ 2ac } ) = \arccos(\dfrac{ 38^2+48.76^2-34^2 }{ 2 \cdot \ 38 \cdot \ 48.76 } ) = 44^\circ \ \\ γ = 180^\circ - α - β = 180^\circ - 50^\circ 55'50" - 44^\circ = 85^\circ 4'10"

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=643.6160.38=10.66T = rs \ \\ r = \dfrac{ T }{ s } = \dfrac{ 643.61 }{ 60.38 } = 10.66

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=38 34 48.764 10.659 60.382=24.47R = \dfrac{ a b c }{ 4 \ r s } = \dfrac{ 38 \cdot \ 34 \cdot \ 48.76 }{ 4 \cdot \ 10.659 \cdot \ 60.382 } = 24.47

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=2 342+2 48.7623822=37.496 mb=2c2+2a2b22=2 48.762+2 3823422=40.273 mc=2a2+2b2c22=2 382+2 34248.7622=26.562m_a = \dfrac{ \sqrt{ 2b^2+2c^2 - a^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 34^2+2 \cdot \ 48.76^2 - 38^2 } }{ 2 } = 37.496 \ \\ m_b = \dfrac{ \sqrt{ 2c^2+2a^2 - b^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 48.76^2+2 \cdot \ 38^2 - 34^2 } }{ 2 } = 40.273 \ \\ m_c = \dfrac{ \sqrt{ 2a^2+2b^2 - c^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 38^2+2 \cdot \ 34^2 - 48.76^2 } }{ 2 } = 26.562



#2 Obtuse scalene triangle.

Sides: a = 38   b = 34   c = 5.90660198465

Area: T = 77.95106563274
Perimeter: p = 77.90660198465
Semiperimeter: s = 38.95330099233

Angle ∠ A = α = 129.0699421568° = 129°4'10″ = 2.253268637 rad
Angle ∠ B = β = 44° = 0.76879448709 rad
Angle ∠ C = γ = 6.93105784325° = 6°55'50″ = 0.12109614127 rad

Height: ha = 4.10326661225
Height: hb = 4.58553327251
Height: hc = 26.39770180774

Median: ma = 15.31114511139
Median: mb = 21.22435844101
Median: mc = 35.9344380924

Inradius: r = 2.00111459058
Circumradius: R = 24.47224611736

Vertex coordinates: A[5.90660198465; 0] B[0; 0] C[27.33549124129; 26.39770180774]
Centroid: CG[11.08803107531; 8.79990060258]
Coordinates of the circumscribed circle: U[2.95330099233; 24.29436429604]
Coordinates of the inscribed circle: I[4.95330099233; 2.00111459058]

Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 50.93105784325° = 50°55'50″ = 2.253268637 rad
∠ B' = β' = 136° = 0.76879448709 rad
∠ C' = γ' = 173.0699421568° = 173°4'10″ = 0.12109614127 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=38 b=34 β=44  b2=a2+c22accosβ 342=382+c22 38 c cos(44)  c254.67c+288=0  p=1;q=54.67;r=288 D=q24pr=54.67241288=1836.78974648 D>0  c1,2=q±D2p=54.67±1836.792 c1,2=27.33491241±21.4288925663 c1=48.7638049792 c2=5.90601984654   Factored form of the equation:  (c48.7638049792)(c5.90601984654)=0   c>0a = 38 \ \\ b = 34 \ \\ β = 44^\circ \ \\ \ \\ b^2 = a^2 + c^2 - 2ac \cos β \ \\ 34^2 = 38^2 + c^2 -2 \cdot \ 38 \cdot \ c \cdot \ \cos (44^\circ ) \ \\ \ \\ c^2 -54.67c +288 =0 \ \\ \ \\ p=1; q=-54.67; r=288 \ \\ D = q^2 - 4pr = 54.67^2 - 4\cdot 1 \cdot 288 = 1836.78974648 \ \\ D>0 \ \\ \ \\ c_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 54.67 \pm \sqrt{ 1836.79 } }{ 2 } \ \\ c_{1,2} = 27.33491241 \pm 21.4288925663 \ \\ c_{1} = 48.7638049792 \ \\ c_{2} = 5.90601984654 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (c -48.7638049792) (c -5.90601984654) = 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=38 b=34 c=5.91a = 38 \ \\ b = 34 \ \\ c = 5.91

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

p=a+b+c=38+34+5.91=77.91p = a+b+c = 38+34+5.91 = 77.91

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=77.912=38.95s = \dfrac{ p }{ 2 } = \dfrac{ 77.91 }{ 2 } = 38.95

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=38.95(38.9538)(38.9534)(38.955.91) T=6076.3=77.95T = \sqrt{ s(s-a)(s-b)(s-c) } \ \\ T = \sqrt{ 38.95(38.95-38)(38.95-34)(38.95-5.91) } \ \\ T = \sqrt{ 6076.3 } = 77.95

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 77.9538=4.1 hb=2 Tb=2 77.9534=4.59 hc=2 Tc=2 77.955.91=26.4T = \dfrac{ a h _a }{ 2 } \ \\ \ \\ h _a = \dfrac{ 2 \ T }{ a } = \dfrac{ 2 \cdot \ 77.95 }{ 38 } = 4.1 \ \\ h _b = \dfrac{ 2 \ T }{ b } = \dfrac{ 2 \cdot \ 77.95 }{ 34 } = 4.59 \ \\ h _c = \dfrac{ 2 \ T }{ c } = \dfrac{ 2 \cdot \ 77.95 }{ 5.91 } = 26.4

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(342+5.9123822 34 5.91)=129410"  b2=a2+c22accosβ β=arccos(a2+c2b22ac)=arccos(382+5.9123422 38 5.91)=44 γ=180αβ=180129410"44=65550"a^2 = b^2+c^2 - 2bc \cos α \ \\ \ \\ α = \arccos(\dfrac{ b^2+c^2-a^2 }{ 2bc } ) = \arccos(\dfrac{ 34^2+5.91^2-38^2 }{ 2 \cdot \ 34 \cdot \ 5.91 } ) = 129^\circ 4'10" \ \\ \ \\ b^2 = a^2+c^2 - 2ac \cos β \ \\ β = \arccos(\dfrac{ a^2+c^2-b^2 }{ 2ac } ) = \arccos(\dfrac{ 38^2+5.91^2-34^2 }{ 2 \cdot \ 38 \cdot \ 5.91 } ) = 44^\circ \ \\ γ = 180^\circ - α - β = 180^\circ - 129^\circ 4'10" - 44^\circ = 6^\circ 55'50"

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=77.9538.95=2T = rs \ \\ r = \dfrac{ T }{ s } = \dfrac{ 77.95 }{ 38.95 } = 2

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=38 34 5.914 2.001 38.953=24.47R = \dfrac{ a b c }{ 4 \ r s } = \dfrac{ 38 \cdot \ 34 \cdot \ 5.91 }{ 4 \cdot \ 2.001 \cdot \ 38.953 } = 24.47

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=2 342+2 5.9123822=15.311 mb=2c2+2a2b22=2 5.912+2 3823422=21.224 mc=2a2+2b2c22=2 382+2 3425.9122=35.934m_a = \dfrac{ \sqrt{ 2b^2+2c^2 - a^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 34^2+2 \cdot \ 5.91^2 - 38^2 } }{ 2 } = 15.311 \ \\ m_b = \dfrac{ \sqrt{ 2c^2+2a^2 - b^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 5.91^2+2 \cdot \ 38^2 - 34^2 } }{ 2 } = 21.224 \ \\ m_c = \dfrac{ \sqrt{ 2a^2+2b^2 - c^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 38^2+2 \cdot \ 34^2 - 5.91^2 } }{ 2 } = 35.934

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