Triangle calculator SSA

Please enter two sides and a non-included angle
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Triangle has two solutions with side c=17.39439811156 and with side c=5.34766770708

#1 Acute scalene triangle.

Sides: a = 47   b = 46   c = 17.39439811156

Area: T = 396.6176680176
Perimeter: p = 110.3943981116
Semiperimeter: s = 55.19769905578

Angle ∠ A = α = 82.47655676591° = 82°28'32″ = 1.43994702081 rad
Angle ∠ B = β = 76° = 1.32664502315 rad
Angle ∠ C = γ = 21.52444323409° = 21°31'28″ = 0.3765672214 rad

Height: ha = 16.87773055394
Height: hb = 17.24442034859
Height: hc = 45.6043899135

Median: ma = 25.63325045504
Median: mb = 26.95987701783
Median: mc = 45.68221885995

Inradius: r = 7.18554765299
Circumradius: R = 23.7044113475

Vertex coordinates: A[17.39439811156; 0] B[0; 0] C[11.37703290932; 45.6043899135]
Centroid: CG[9.58881034029; 15.20112997117]
Coordinates of the circumscribed circle: U[8.69769905578; 22.05110170032]
Coordinates of the inscribed circle: I[9.19769905578; 7.18554765299]

Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 97.52444323409° = 97°31'28″ = 1.43994702081 rad
∠ B' = β' = 104° = 1.32664502315 rad
∠ C' = γ' = 158.4765567659° = 158°28'32″ = 0.3765672214 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=47 b=46 β=76  b2=a2+c22accosβ 462=472+c22 47 c cos(76)  c222.741c+93=0  p=1;q=22.741;r=93 D=q24pr=22.74124193=145.137534749 D>0  c1,2=q±D2p=22.74±145.142 c1,2=11.37032909±6.02365202243 c1=17.3939811156 c2=5.34667707076   Factored form of the equation:  (c17.3939811156)(c5.34667707076)=0   c>0a = 47 \ \\ b = 46 \ \\ β = 76^\circ \ \\ \ \\ b^2 = a^2 + c^2 - 2ac \cos β \ \\ 46^2 = 47^2 + c^2 -2 \cdot \ 47 \cdot \ c \cdot \ \cos (76^\circ ) \ \\ \ \\ c^2 -22.741c +93 =0 \ \\ \ \\ p=1; q=-22.741; r=93 \ \\ D = q^2 - 4pr = 22.741^2 - 4\cdot 1 \cdot 93 = 145.137534749 \ \\ D>0 \ \\ \ \\ c_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 22.74 \pm \sqrt{ 145.14 } }{ 2 } \ \\ c_{1,2} = 11.37032909 \pm 6.02365202243 \ \\ c_{1} = 17.3939811156 \ \\ c_{2} = 5.34667707076 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (c -17.3939811156) (c -5.34667707076) = 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=47 b=46 c=17.39a = 47 \ \\ b = 46 \ \\ c = 17.39

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

p=a+b+c=47+46+17.39=110.39p = a+b+c = 47+46+17.39 = 110.39

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=110.392=55.2s = \dfrac{ p }{ 2 } = \dfrac{ 110.39 }{ 2 } = 55.2

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=55.2(55.247)(55.246)(55.217.39) T=157304.79=396.62T = \sqrt{ s(s-a)(s-b)(s-c) } \ \\ T = \sqrt{ 55.2(55.2-47)(55.2-46)(55.2-17.39) } \ \\ T = \sqrt{ 157304.79 } = 396.62

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=2 396.6247=16.88 hb=2 Tb=2 396.6246=17.24 hc=2 Tc=2 396.6217.39=45.6T = \dfrac{ a h _a }{ 2 } \ \\ \ \\ h _a = \dfrac{ 2 \ T }{ a } = \dfrac{ 2 \cdot \ 396.62 }{ 47 } = 16.88 \ \\ h _b = \dfrac{ 2 \ T }{ b } = \dfrac{ 2 \cdot \ 396.62 }{ 46 } = 17.24 \ \\ h _c = \dfrac{ 2 \ T }{ c } = \dfrac{ 2 \cdot \ 396.62 }{ 17.39 } = 45.6

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(462+17.3924722 46 17.39)=822832"  b2=a2+c22accosβ β=arccos(a2+c2b22ac)=arccos(472+17.3924622 47 17.39)=76 γ=180αβ=180822832"76=213128"a^2 = b^2+c^2 - 2bc \cos α \ \\ \ \\ α = \arccos(\dfrac{ b^2+c^2-a^2 }{ 2bc } ) = \arccos(\dfrac{ 46^2+17.39^2-47^2 }{ 2 \cdot \ 46 \cdot \ 17.39 } ) = 82^\circ 28'32" \ \\ \ \\ b^2 = a^2+c^2 - 2ac \cos β \ \\ β = \arccos(\dfrac{ a^2+c^2-b^2 }{ 2ac } ) = \arccos(\dfrac{ 47^2+17.39^2-46^2 }{ 2 \cdot \ 47 \cdot \ 17.39 } ) = 76^\circ \ \\ γ = 180^\circ - α - β = 180^\circ - 82^\circ 28'32" - 76^\circ = 21^\circ 31'28"

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=396.6255.2=7.19T = rs \ \\ r = \dfrac{ T }{ s } = \dfrac{ 396.62 }{ 55.2 } = 7.19

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=47 46 17.394 7.185 55.197=23.7R = \dfrac{ a b c }{ 4 \ r s } = \dfrac{ 47 \cdot \ 46 \cdot \ 17.39 }{ 4 \cdot \ 7.185 \cdot \ 55.197 } = 23.7

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=2 462+2 17.3924722=25.633 mb=2c2+2a2b22=2 17.392+2 4724622=26.959 mc=2a2+2b2c22=2 472+2 46217.3922=45.682m_a = \dfrac{ \sqrt{ 2b^2+2c^2 - a^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 46^2+2 \cdot \ 17.39^2 - 47^2 } }{ 2 } = 25.633 \ \\ m_b = \dfrac{ \sqrt{ 2c^2+2a^2 - b^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 17.39^2+2 \cdot \ 47^2 - 46^2 } }{ 2 } = 26.959 \ \\ m_c = \dfrac{ \sqrt{ 2a^2+2b^2 - c^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 47^2+2 \cdot \ 46^2 - 17.39^2 } }{ 2 } = 45.682



#2 Obtuse scalene triangle.

Sides: a = 47   b = 46   c = 5.34766770708

Area: T = 121.9154660921
Perimeter: p = 98.34766770708
Semiperimeter: s = 49.17333385354

Angle ∠ A = α = 97.52444323409° = 97°31'28″ = 1.70221224455 rad
Angle ∠ B = β = 76° = 1.32664502315 rad
Angle ∠ C = γ = 6.47655676591° = 6°28'32″ = 0.11330199766 rad

Height: ha = 5.18878579115
Height: hb = 5.30106374313
Height: hc = 45.6043899135

Median: ma = 22.8044461797
Median: mb = 24.28656640397
Median: mc = 46.42657822882

Inradius: r = 2.47992837857
Circumradius: R = 23.7044113475

Vertex coordinates: A[5.34766770708; 0] B[0; 0] C[11.37703290932; 45.6043899135]
Centroid: CG[5.5722335388; 15.20112997117]
Coordinates of the circumscribed circle: U[2.67333385354; 23.55328821318]
Coordinates of the inscribed circle: I[3.17333385354; 2.47992837857]

Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 82.47655676591° = 82°28'32″ = 1.70221224455 rad
∠ B' = β' = 104° = 1.32664502315 rad
∠ C' = γ' = 173.5244432341° = 173°31'28″ = 0.11330199766 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=47 b=46 β=76  b2=a2+c22accosβ 462=472+c22 47 c cos(76)  c222.741c+93=0  p=1;q=22.741;r=93 D=q24pr=22.74124193=145.137534749 D>0  c1,2=q±D2p=22.74±145.142 c1,2=11.37032909±6.02365202243 c1=17.3939811156 c2=5.34667707076   Factored form of the equation:  (c17.3939811156)(c5.34667707076)=0   c>0a = 47 \ \\ b = 46 \ \\ β = 76^\circ \ \\ \ \\ b^2 = a^2 + c^2 - 2ac \cos β \ \\ 46^2 = 47^2 + c^2 -2 \cdot \ 47 \cdot \ c \cdot \ \cos (76^\circ ) \ \\ \ \\ c^2 -22.741c +93 =0 \ \\ \ \\ p=1; q=-22.741; r=93 \ \\ D = q^2 - 4pr = 22.741^2 - 4\cdot 1 \cdot 93 = 145.137534749 \ \\ D>0 \ \\ \ \\ c_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 22.74 \pm \sqrt{ 145.14 } }{ 2 } \ \\ c_{1,2} = 11.37032909 \pm 6.02365202243 \ \\ c_{1} = 17.3939811156 \ \\ c_{2} = 5.34667707076 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (c -17.3939811156) (c -5.34667707076) = 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=47 b=46 c=5.35a = 47 \ \\ b = 46 \ \\ c = 5.35

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

p=a+b+c=47+46+5.35=98.35p = a+b+c = 47+46+5.35 = 98.35

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=98.352=49.17s = \dfrac{ p }{ 2 } = \dfrac{ 98.35 }{ 2 } = 49.17

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=49.17(49.1747)(49.1746)(49.175.35) T=14863.18=121.91T = \sqrt{ s(s-a)(s-b)(s-c) } \ \\ T = \sqrt{ 49.17(49.17-47)(49.17-46)(49.17-5.35) } \ \\ T = \sqrt{ 14863.18 } = 121.91

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=2 121.9147=5.19 hb=2 Tb=2 121.9146=5.3 hc=2 Tc=2 121.915.35=45.6T = \dfrac{ a h _a }{ 2 } \ \\ \ \\ h _a = \dfrac{ 2 \ T }{ a } = \dfrac{ 2 \cdot \ 121.91 }{ 47 } = 5.19 \ \\ h _b = \dfrac{ 2 \ T }{ b } = \dfrac{ 2 \cdot \ 121.91 }{ 46 } = 5.3 \ \\ h _c = \dfrac{ 2 \ T }{ c } = \dfrac{ 2 \cdot \ 121.91 }{ 5.35 } = 45.6

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(462+5.3524722 46 5.35)=973128"  b2=a2+c22accosβ β=arccos(a2+c2b22ac)=arccos(472+5.3524622 47 5.35)=76 γ=180αβ=180973128"76=62832"a^2 = b^2+c^2 - 2bc \cos α \ \\ \ \\ α = \arccos(\dfrac{ b^2+c^2-a^2 }{ 2bc } ) = \arccos(\dfrac{ 46^2+5.35^2-47^2 }{ 2 \cdot \ 46 \cdot \ 5.35 } ) = 97^\circ 31'28" \ \\ \ \\ b^2 = a^2+c^2 - 2ac \cos β \ \\ β = \arccos(\dfrac{ a^2+c^2-b^2 }{ 2ac } ) = \arccos(\dfrac{ 47^2+5.35^2-46^2 }{ 2 \cdot \ 47 \cdot \ 5.35 } ) = 76^\circ \ \\ γ = 180^\circ - α - β = 180^\circ - 97^\circ 31'28" - 76^\circ = 6^\circ 28'32"

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=121.9149.17=2.48T = rs \ \\ r = \dfrac{ T }{ s } = \dfrac{ 121.91 }{ 49.17 } = 2.48

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=47 46 5.354 2.479 49.173=23.7R = \dfrac{ a b c }{ 4 \ r s } = \dfrac{ 47 \cdot \ 46 \cdot \ 5.35 }{ 4 \cdot \ 2.479 \cdot \ 49.173 } = 23.7

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=2 462+2 5.3524722=22.804 mb=2c2+2a2b22=2 5.352+2 4724622=24.286 mc=2a2+2b2c22=2 472+2 4625.3522=46.426m_a = \dfrac{ \sqrt{ 2b^2+2c^2 - a^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 46^2+2 \cdot \ 5.35^2 - 47^2 } }{ 2 } = 22.804 \ \\ m_b = \dfrac{ \sqrt{ 2c^2+2a^2 - b^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 5.35^2+2 \cdot \ 47^2 - 46^2 } }{ 2 } = 24.286 \ \\ m_c = \dfrac{ \sqrt{ 2a^2+2b^2 - c^2 } }{ 2 } = \dfrac{ \sqrt{ 2 \cdot \ 47^2+2 \cdot \ 46^2 - 5.35^2 } }{ 2 } = 46.426

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