Triangle calculator SSA

Please enter two sides and a non-included angle
°


Triangle has two solutions with side c=59.77992384739 and with side c=22.88441991789

#1 Acute scalene triangle.

Sides: a = 63   b = 51   c = 59.77992384739

Area: T = 1421.15328652
Perimeter: p = 173.7799238474
Semiperimeter: s = 86.89896192369

Angle ∠ A = α = 68.79443766898° = 68°47'40″ = 1.20106883801 rad
Angle ∠ B = β = 49° = 0.85552113335 rad
Angle ∠ C = γ = 62.20656233102° = 62°12'20″ = 1.086569294 rad

Height: ha = 45.11659639747
Height: hb = 55.73114849099
Height: hc = 47.5476703554

Median: ma = 45.77114832211
Median: mb = 55.86661675458
Median: mc = 48.90440965755

Inradius: r = 16.35658417873
Circumradius: R = 33.78878313304

Vertex coordinates: A[59.77992384739; 0] B[0; 0] C[41.33217188264; 47.5476703554]
Centroid: CG[33.70436524334; 15.84989011847]
Coordinates of the circumscribed circle: U[29.89896192369; 15.75552596894]
Coordinates of the inscribed circle: I[35.89896192369; 16.35658417873]

Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 111.206562331° = 111°12'20″ = 1.20106883801 rad
∠ B' = β' = 131° = 0.85552113335 rad
∠ C' = γ' = 117.794437669° = 117°47'40″ = 1.086569294 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=63 b=51 β=49  b2=a2+c22accosβ 512=632+c22 63 c cos(49)  c282.663c+1368=0  p=1;q=82.663;r=1368 D=q24pr=82.6632411368=1361.24392458 D>0  c1,2=q±D2p=82.66±1361.242 c1,2=41.33171883±18.4475196475 c1=59.7792384739 c2=22.8841991789   Factored form of the equation:  (c59.7792384739)(c22.8841991789)=0  c>0a = 63 \ \\ b = 51 \ \\ β = 49^\circ \ \\ \ \\ b^2 = a^2 + c^2 - 2ac \cos β \ \\ 51^2 = 63^2 + c^2 -2 \cdot \ 63 \cdot \ c \cdot \ \cos (49^\circ ) \ \\ \ \\ c^2 -82.663c +1368 =0 \ \\ \ \\ p=1; q=-82.663; r=1368 \ \\ D = q^2 - 4pr = 82.663^2 - 4\cdot 1 \cdot 1368 = 1361.24392458 \ \\ D>0 \ \\ \ \\ c_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 82.66 \pm \sqrt{ 1361.24 } }{ 2 } \ \\ c_{1,2} = 41.33171883 \pm 18.4475196475 \ \\ c_{1} = 59.7792384739 \ \\ c_{2} = 22.8841991789 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (c -59.7792384739) (c -22.8841991789) = 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=63 b=51 c=59.78a = 63 \ \\ b = 51 \ \\ c = 59.78

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

p=a+b+c=63+51+59.78=173.78p = a+b+c = 63+51+59.78 = 173.78

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=173.782=86.89s = \dfrac{ p }{ 2 } = \dfrac{ 173.78 }{ 2 } = 86.89

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=86.89(86.8963)(86.8951)(86.8959.78) T=2019675.47=1421.15T = \sqrt{ s(s-a)(s-b)(s-c) } \ \\ T = \sqrt{ 86.89(86.89-63)(86.89-51)(86.89-59.78) } \ \\ T = \sqrt{ 2019675.47 } = 1421.15

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 1421.1563=45.12 hb=2 Tb=2 1421.1551=55.73 hc=2 Tc=2 1421.1559.78=47.55T = \dfrac{ a h _a }{ 2 } \ \\ \ \\ h _a = \dfrac{ 2 \ T }{ a } = \dfrac{ 2 \cdot \ 1421.15 }{ 63 } = 45.12 \ \\ h _b = \dfrac{ 2 \ T }{ b } = \dfrac{ 2 \cdot \ 1421.15 }{ 51 } = 55.73 \ \\ h _c = \dfrac{ 2 \ T }{ c } = \dfrac{ 2 \cdot \ 1421.15 }{ 59.78 } = 47.55

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(512+59.7826322 51 59.78)=684740"  b2=a2+c22accosβ β=arccos(a2+c2b22ac)=arccos(632+59.7825122 63 59.78)=49 γ=180αβ=180684740"49=621220"a^2 = b^2+c^2 - 2bc \cos α \ \\ \ \\ α = \arccos(\dfrac{ b^2+c^2-a^2 }{ 2bc } ) = \arccos(\dfrac{ 51^2+59.78^2-63^2 }{ 2 \cdot \ 51 \cdot \ 59.78 } ) = 68^\circ 47'40" \ \\ \ \\ b^2 = a^2+c^2 - 2ac \cos β \ \\ β = \arccos(\dfrac{ a^2+c^2-b^2 }{ 2ac } ) = \arccos(\dfrac{ 63^2+59.78^2-51^2 }{ 2 \cdot \ 63 \cdot \ 59.78 } ) = 49^\circ \ \\ γ = 180^\circ - α - β = 180^\circ - 68^\circ 47'40" - 49^\circ = 62^\circ 12'20"

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=1421.1586.89=16.36T = rs \ \\ r = \dfrac{ T }{ s } = \dfrac{ 1421.15 }{ 86.89 } = 16.36

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=63 51 59.784 16.356 86.89=33.79R = \dfrac{ a b c }{ 4 \ r s } = \dfrac{ 63 \cdot \ 51 \cdot \ 59.78 }{ 4 \cdot \ 16.356 \cdot \ 86.89 } = 33.79

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=2 512+2 59.7826322=45.771 mb=2c2+2a2b22=2 59.782+2 6325122=55.866 mc=2a2+2b2c22=2 632+2 51259.7822=48.904m_a = \dfrac{ \sqrt{ 2b^2+2c^2 - a^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 51^2+2 \cdot \ 59.78^2 - 63^2 } }{ 2 } = 45.771 \ \\ m_b = \dfrac{ \sqrt{ 2c^2+2a^2 - b^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 59.78^2+2 \cdot \ 63^2 - 51^2 } }{ 2 } = 55.866 \ \\ m_c = \dfrac{ \sqrt{ 2a^2+2b^2 - c^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 63^2+2 \cdot \ 51^2 - 59.78^2 } }{ 2 } = 48.904



#2 Obtuse scalene triangle.

Sides: a = 63   b = 51   c = 22.88441991789

Area: T = 544.0344117216
Perimeter: p = 136.8844199179
Semiperimeter: s = 68.44220995895

Angle ∠ A = α = 111.206562331° = 111°12'20″ = 1.94109042735 rad
Angle ∠ B = β = 49° = 0.85552113335 rad
Angle ∠ C = γ = 19.79443766898° = 19°47'40″ = 0.34554770466 rad

Height: ha = 17.2710924356
Height: hb = 21.33546712634
Height: hc = 47.5476703554

Median: ma = 23.87766263536
Median: mb = 39.95111362295
Median: mc = 56.16111819408

Inradius: r = 7.94988227345
Circumradius: R = 33.78878313304

Vertex coordinates: A[22.88441991789; 0] B[0; 0] C[41.33217188264; 47.5476703554]
Centroid: CG[21.40553060018; 15.84989011847]
Coordinates of the circumscribed circle: U[11.44220995895; 31.79114438646]
Coordinates of the inscribed circle: I[17.44220995895; 7.94988227345]

Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 68.79443766898° = 68°47'40″ = 1.94109042735 rad
∠ B' = β' = 131° = 0.85552113335 rad
∠ C' = γ' = 160.206562331° = 160°12'20″ = 0.34554770466 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=63 b=51 β=49  b2=a2+c22accosβ 512=632+c22 63 c cos(49)  c282.663c+1368=0  p=1;q=82.663;r=1368 D=q24pr=82.6632411368=1361.24392458 D>0  c1,2=q±D2p=82.66±1361.242 c1,2=41.33171883±18.4475196475 c1=59.7792384739 c2=22.8841991789   Factored form of the equation:  (c59.7792384739)(c22.8841991789)=0  c>0a = 63 \ \\ b = 51 \ \\ β = 49^\circ \ \\ \ \\ b^2 = a^2 + c^2 - 2ac \cos β \ \\ 51^2 = 63^2 + c^2 -2 \cdot \ 63 \cdot \ c \cdot \ \cos (49^\circ ) \ \\ \ \\ c^2 -82.663c +1368 =0 \ \\ \ \\ p=1; q=-82.663; r=1368 \ \\ D = q^2 - 4pr = 82.663^2 - 4\cdot 1 \cdot 1368 = 1361.24392458 \ \\ D>0 \ \\ \ \\ c_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 82.66 \pm \sqrt{ 1361.24 } }{ 2 } \ \\ c_{1,2} = 41.33171883 \pm 18.4475196475 \ \\ c_{1} = 59.7792384739 \ \\ c_{2} = 22.8841991789 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (c -59.7792384739) (c -22.8841991789) = 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=63 b=51 c=22.88a = 63 \ \\ b = 51 \ \\ c = 22.88

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

p=a+b+c=63+51+22.88=136.88p = a+b+c = 63+51+22.88 = 136.88

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=136.882=68.44s = \dfrac{ p }{ 2 } = \dfrac{ 136.88 }{ 2 } = 68.44

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=68.44(68.4463)(68.4451)(68.4422.88) T=295973.12=544.03T = \sqrt{ s(s-a)(s-b)(s-c) } \ \\ T = \sqrt{ 68.44(68.44-63)(68.44-51)(68.44-22.88) } \ \\ T = \sqrt{ 295973.12 } = 544.03

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=2 544.0363=17.27 hb=2 Tb=2 544.0351=21.33 hc=2 Tc=2 544.0322.88=47.55T = \dfrac{ a h _a }{ 2 } \ \\ \ \\ h _a = \dfrac{ 2 \ T }{ a } = \dfrac{ 2 \cdot \ 544.03 }{ 63 } = 17.27 \ \\ h _b = \dfrac{ 2 \ T }{ b } = \dfrac{ 2 \cdot \ 544.03 }{ 51 } = 21.33 \ \\ h _c = \dfrac{ 2 \ T }{ c } = \dfrac{ 2 \cdot \ 544.03 }{ 22.88 } = 47.55

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(512+22.8826322 51 22.88)=1111220"  b2=a2+c22accosβ β=arccos(a2+c2b22ac)=arccos(632+22.8825122 63 22.88)=49 γ=180αβ=1801111220"49=194740"a^2 = b^2+c^2 - 2bc \cos α \ \\ \ \\ α = \arccos(\dfrac{ b^2+c^2-a^2 }{ 2bc } ) = \arccos(\dfrac{ 51^2+22.88^2-63^2 }{ 2 \cdot \ 51 \cdot \ 22.88 } ) = 111^\circ 12'20" \ \\ \ \\ b^2 = a^2+c^2 - 2ac \cos β \ \\ β = \arccos(\dfrac{ a^2+c^2-b^2 }{ 2ac } ) = \arccos(\dfrac{ 63^2+22.88^2-51^2 }{ 2 \cdot \ 63 \cdot \ 22.88 } ) = 49^\circ \ \\ γ = 180^\circ - α - β = 180^\circ - 111^\circ 12'20" - 49^\circ = 19^\circ 47'40"

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=544.0368.44=7.95T = rs \ \\ r = \dfrac{ T }{ s } = \dfrac{ 544.03 }{ 68.44 } = 7.95

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=63 51 22.884 7.949 68.442=33.79R = \dfrac{ a b c }{ 4 \ r s } = \dfrac{ 63 \cdot \ 51 \cdot \ 22.88 }{ 4 \cdot \ 7.949 \cdot \ 68.442 } = 33.79

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=2 512+2 22.8826322=23.877 mb=2c2+2a2b22=2 22.882+2 6325122=39.951 mc=2a2+2b2c22=2 632+2 51222.8822=56.161m_a = \dfrac{ \sqrt{ 2b^2+2c^2 - a^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 51^2+2 \cdot \ 22.88^2 - 63^2 } }{ 2 } = 23.877 \ \\ m_b = \dfrac{ \sqrt{ 2c^2+2a^2 - b^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 22.88^2+2 \cdot \ 63^2 - 51^2 } }{ 2 } = 39.951 \ \\ m_c = \dfrac{ \sqrt{ 2a^2+2b^2 - c^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 63^2+2 \cdot \ 51^2 - 22.88^2 } }{ 2 } = 56.161

Calculate another triangle


Look also our friend's collection of math problems and questions:

See more information about triangles or more details on solving triangles.