Triangle Calculator (Solve for Sides, Angles, Area and More)

Enter the three values you know out of the sides a, b, c and the angles A, B, C (in degrees). At least one must be a side. Side a is across from angle A, side b from angle B, and side c from angle C.

Side a (opposite angle A)

Angle A (degrees)

Side b (opposite angle B)

Angle B (degrees)

Side c (opposite angle C)

Angle C (degrees)

Fill in exactly three fields, at least one of them a side. Enter angles in degrees (°), for example 90 for a right angle. Leave the other fields blank.

Result and figure

Enter the three values you know out of the three sides and three angles (at least one must be a side) in the fields on the left and press "Calculate". The result will appear here.

[ CLOSE ]

What you can do on this page
What is this calculation used for?
How to Use
Formulas and figures
Symbols and terms
- Symbols
- Terms
Good to know before you start
How to calculate it in Excel
How to calculate it in Google Sheets
How to calculate it in Python
How to write it in LaTeX and other math languages (copy and paste)
How to have ChatGPT do the calculation
DataChef Features
Related Features
NumberChef Calculators List

What you can do on this page

Enter any three of the sides \(a, b, c\) and the angles \(A, B, C\) that you know (at least one must be a side), and get all the other sides and angles
Also finds the area, the perimeter, the semiperimeter, the three heights, the three medians, the inradius and the circumradius, all at once
Automatically tells you the type of triangle: equilateral, isosceles or scalene, and acute, right or obtuse
When two sides and an angle opposite one of them (SSA) allow two different triangles, both solutions are shown (the well-known "ambiguous case")
A plain-language explanation of the law of cosines, the law of sines and Heron's formula, and copy-and-paste formulas for Excel, Google Sheets and Python are all on this page

Sides and angles are paired by letter: side a is across from angle A (its opposite side), and so on, with lowercase for sides and uppercase for angles. Enter angles in degrees (°).

What is this calculation used for?

Surveying and mapmaking (measuring the distance to a place you cannot reach)

You can find the distance to a landmark across a river without stretching a tape measure over the water. Lay out a 300-foot baseline on your side and measure the angle to the landmark from each end. That gives a triangle with one side (300 feet) and two angles. If the angles are 50° and 60°, the third angle is 70°, and the law of sines gives the distance to the landmark as \(300 \times \sin 50^\circ \div \sin 70^\circ \approx 244.6\) feet.
The total stations that surveyors use, and the triangulation that mapmaking was built on, come down to repeating this very calculation.

Measuring the area of a triangular lot or field (Heron's formula)

For a triangular piece of land, you can find the area with just a tape measure, without any tool for measuring angles. If the three sides are 90 feet, 120 feet and 150 feet, the semiperimeter is 180 feet and the area is \(\sqrt{180 \times 90 \times 60 \times 30} = 5{,}400\) square feet (about 0.12 acre).
"Anyone can measure lengths accurately" is why Heron's formula has been popular in practice, and it is still handy for a rough check of the area of farmland or a building lot.

Construction and DIY (working out the length and angle of slanted parts first)

Buildings are full of slanted parts: roof slopes, stairs, diagonal braces and so on. For a shed roof that runs 12 feet across and rises 5 feet, the length of the slope is \(\sqrt{12^2 + 5^2} = 13\) feet (with a right angle, the correction term of the law of cosines disappears, leaving the Pythagorean theorem).
If you work out the lengths and angles before cutting the lumber, you waste less material and redo less work on site.

Robot arms and machine design (two links and the joint angle)

A factory robot arm is itself a triangle: two arm links and the joint angle between them. With links of 20 inches and 12 inches and the joint opened to 120°, the law of cosines gives the distance from the base to the hand as \(\sqrt{20^2 + 12^2 - 2 \times 20 \times 12 \times \cos 120^\circ} = 28\) inches.
Working backward to the joint angles that move the hand to a target position (inverse kinematics) also uses a rearranged law of cosines as a basic building block.

Astronomy (measuring the distance to stars with triangles)

The distance to a star is measured with a very long, thin triangle whose base is the distance the Earth moves in half a year of its orbit around the Sun (this is called stellar parallax). Finding the distance from the base and the small shift in angle is the same idea as this page's "find the rest from one side and angles" calculation.
"If you can solve a triangle, you can find the distance to a place you cannot go." This works from land surveying all the way to outer space, and it is one of the most powerful uses of geometry.

Formulas and figures

Angle sum of a triangle

Figure

Standard notation (the usual math form)

\(A\) \(+\) \(B\) \(+\) \(C\) \(=\) \(180^\circ\)

In words (symbols replaced with words)

① \(A\): angle A \(+\) ② \(B\): angle B \(+\) ③ \(C\): angle C \(=\) ④ \(180^\circ\) (always)

The formula in words

① Add angle \(A\)

② plus angle \(B\)

③ plus angle \(C\)

④ and you always get \(180^\circ\)

Quick example

In a triangle with two angles of 50° and 60°, the third angle is

third angle \(C\) \(=\) \(180^\circ\) \(-\) angle \(A\) (50°) \(-\) angle \(B\) (60°)

\(180^\circ - 50^\circ - 60^\circ = 70^\circ\)

Key idea

In every triangle, whatever its shape, the three angles add up to 180°. So once you know two angles, the third is just a subtraction. For "one side and two angles" input, this calculator uses this formula first to find the third angle.

Law of cosines (third side from two sides and the included angle)

Figure

Standard notation (the usual math form)

\(c^{2}\) \(=\) \(a^{2}\) \(+\) \(b^{2}\) \(-\) \(2ab\cos C\)

In words (symbols replaced with words)

④ \(c^{2}\): the third side \(c\), squared \(=\) ① \(a^{2}\): the known side \(a\), squared \(+\) ② \(b^{2}\): the known side \(b\), squared \(-\) ③ \(2ab\cos C\): correction term

The formula in words

① Take the known side \(a\), squared

② add the known side \(b\), squared

③ subtract the correction term \(2ab\cos C\), set by how wide the included angle is

④ and you get the third side \(c\), squared (take the square root at the end)

Quick example

With two sides of 5 and 7 and an included angle of 60° (cos 60° = 0.5), the third side is

third side \(c\), squared \(=\) side \(a\) squared (5²) \(+\) side \(b\) squared (7²) \(-\) correction term (2×5×7×0.5)

\(c^{2} = 5^{2} + 7^{2} - 2 \times 5 \times 7 \times \cos 60^\circ = 25 + 49 - 35 = 39\)

\(c = \sqrt{39} \approx 6.245\)

Key idea

\(\cos C\) (cosine) expresses how wide the included angle is as a number from −1 to 1. When the angle is exactly 90°, \(\cos 90^\circ = 0\), the correction term disappears, and you get the Pythagorean theorem \(c^2 = a^2 + b^2\). In other words, the law of cosines is the Pythagorean theorem extended to triangles without a right angle. When you know all three sides and want an angle, use the rearranged form \(A = \arccos\bigl(\frac{b^2 + c^2 - a^2}{2bc}\bigr)\) (the "three sides" input of this calculator uses this form).

Law of sines (the bridge between sides and opposite angles)

Figure

Standard notation (the usual math form)

\(\dfrac{a}{\sin A}\) \(=\) \(\dfrac{b}{\sin B}\) \(=\) \(\dfrac{c}{\sin C}\) \(=\) \(2R\)

In words (symbols replaced with words)

① \(\dfrac{a}{\sin A}\): side \(a\) ÷ sine of its opposite angle \(=\) ② \(\dfrac{b}{\sin B}\): side \(b\) ÷ sine of its opposite angle \(=\) ③ \(\dfrac{c}{\sin C}\): side \(c\) ÷ sine of its opposite angle \(=\) ④ \(2R\): diameter of the circumscribed circle

The formula in words

① Side \(a\) divided by the sine of its opposite angle \(\dfrac{a}{\sin A}\)

② side \(b\) divided by the sine of its opposite angle \(\dfrac{b}{\sin B}\)

③ and side \(c\) divided by the sine of its opposite angle \(\dfrac{c}{\sin C}\) all have the same value,

④ which equals the diameter of the circumscribed circle \(2R\)

Quick example

With side a = 6, its opposite angle A = 45° and angle B = 60°, side b is

side \(b\) \(=\) side \(a\) (6) \(\times\) sine of the opposite angle \(\sin 60^\circ\) \(\div\) sine of the opposite angle \(\sin 45^\circ\)

\(b = \dfrac{6 \times \sin 60^\circ}{\sin 45^\circ} \approx 7.348\)

Key idea

"Side ÷ sine of its opposite angle" gives the same value for all three pairs. So if you know one pair (a side and its opposite angle), you can use it as a bridge to find the other sides and angles one after another. The "one side and two angles" input of this calculator uses this formula. On top of that, this common value is equal to the diameter \(2R\) of the circle through the three corners of the triangle (the circumscribed circle). The full law of sines includes this fact too.

Area (two sides and the included angle)

Figure

Standard notation (the usual math form)

\(S\) \(=\) \(\dfrac{1}{2}\) \(\times\) \(a\) \(\times\) \(b\) \(\times\) \(\sin C\)

In words (symbols replaced with words)

⑤ \(S\): area \(=\) ④ one half \(\times\) ① \(a\): known side \(\times\) ② \(b\): known side \(\times\) ③ \(\sin C\): sine of the included angle

The formula in words

① Multiply the known side \(a\)

② by the known side \(b\)

③ and by the sine of the included angle \(\sin C\)

④ then take one half of it

⑤ and you get the area \(S\)

Quick example

The area of a triangle with two sides of 8 and 5 and an included angle of 30° (sin 30° = 0.5) is

area \(S\) \(=\) one half \(\times\) side \(a\) (8) \(\times\) side \(b\) (5) \(\times\) sine of the included angle (0.5)

\(S = \dfrac{1}{2} \times 8 \times 5 \times \sin 30^\circ = \dfrac{1}{2} \times 8 \times 5 \times 0.5 = 10\)

Key idea

This is the familiar "base × height ÷ 2" from elementary school, with the height replaced by \(b\sin C\) (the height worked out from side b and the included angle). When the included angle is 90°, \(\sin 90^\circ = 1\), so the height is side b itself and the area is as large as it can be.

Heron's formula (area from three sides)

Figure

Standard notation (the usual math form)

\(s\) \(=\) \(\dfrac{a+b+c}{2}\)

\(S\) \(=\) \(\sqrt{s(s-a)(s-b)(s-c)}\)

In words (symbols replaced with words)

② \(s\): semiperimeter \(=\) ① sum of the three sides ÷ 2

④ \(S\): area \(=\) ③ square root of \(s\) times "\(s\) − each side" (three factors)

The formula in words

① Add up all three sides \(a, b, c\) and divide by 2. \(\dfrac{a+b+c}{2}\)

② Call this the semiperimeter \(s\)

③ Next, multiply \(s\) by the three numbers you get by subtracting each side from \(s\) (\(s-a\), \(s-b\), \(s-c\)) and take the square root

④ and the result is the area \(S\)

Quick example

The area of a triangle with sides 3, 4 and 5 is

semiperimeter \(s\) \(=\) sum of the three sides (3+4+5) ÷ 2

\(s = \dfrac{3+4+5}{2} = 6\)

\(S = \sqrt{6 \times (6-3) \times (6-4) \times (6-5)} = \sqrt{6 \times 3 \times 2 \times 1} = \sqrt{36} = 6\)

Key idea

The strength of Heron's formula is that it gives the area from the three side lengths alone, without measuring any angle or height. Measuring three sides with a tape measure is enough, so it has long been used for things like finding the area of a triangular plot of land. The trick is to work out the semiperimeter \(s\) (half the perimeter) first.

Radius of the inscribed circle (inradius)

Figure

Standard notation (the usual math form)

In words (symbols replaced with words)

\(r\) \(=\) \(S\) \(\div\) \(s\)

③ \(r\): inradius \(=\) ① \(S\): area \(\div\) ② \(s\): semiperimeter

The formula in words

① Take the area \(S\)

② divide it by the semiperimeter \(s\) (half the perimeter)

③ and you get the inradius \(r\)

Quick example

For the triangle with sides 3, 4 and 5 (area 6, semiperimeter 6), the inradius is

inradius \(r\) \(=\) area (6) \(\div\) semiperimeter (6)

\(r = 6 \div 6 = 1\)

Key idea

The inscribed circle (incircle) is the circle that touches all three sides of the triangle from the inside. If you cut the triangle into three small triangles from the center of the incircle (the incenter), each has "base = one side, height = r", so the total area is \(S = \frac{1}{2}r(a+b+c) = rs\). Solving for \(r\) gives \(r = S \div s\).

Radius of the circumscribed circle (circumradius)

Figure

Standard notation (the usual math form)

In words (symbols replaced with words)

\(R\) \(=\) \(abc\) \(\div\) \(4S\)

③ \(R\): circumradius \(=\) ① \(abc\): product of the three sides \(\div\) ② \(4S\): four times the area

The formula in words

① Take the product of all three sides \(abc\)

② divide it by four times the area \(S\)

③ and you get the circumradius \(R\)

Quick example

For the triangle with sides 3, 4 and 5 (area 6), the circumradius is

circumradius \(R\) \(=\) product of the three sides (3×4×5) \(\div\) four times the area (4×6)

\(R = \dfrac{3 \times 4 \times 5}{4 \times 6} = \dfrac{60}{24} = 2.5\)

Key idea

The circumscribed circle (circumcircle) is the circle through all three corners (vertices) of the triangle. This formula comes from combining the law of sines \(\frac{a}{\sin A} = 2R\) with the area formula \(S = \frac{1}{2}bc\sin A\). The 3-4-5 triangle in the example is a right triangle, so the center of its circumcircle is exactly at the midpoint of the hypotenuse, and \(R\) is half the hypotenuse (\(5 \div 2 = 2.5\)). The result matches exactly.

A triangle is fixed by three pieces of information (at least one of them a side). For three sides, use the law of cosines solved for the angle. For two sides and the included angle, use the law of cosines. For one side and two angles, use the angle sum and the law of sines. These three tools give all the other sides and angles, and from there the area, the incircle and the circumcircle follow one after another.

Symbols and terms

Symbols

\(a, b, c\)	a, b, c (lowercase)	The lengths of the three sides of the triangle. By custom, each side uses the lowercase of the letter of the angle across from it (side a is opposite angle A).
\(A, B, C\)	A, B, C (uppercase)	The three angles of the triangle (in degrees). They always add up to 180°.
\(S\)	S	The area of the triangle. (Some US textbooks use \(K\) or \(A\) for area instead.)
\(s\)	s (lowercase)	The semiperimeter, half the perimeter (\(a+b+c\)). It is used in Heron's formula and in the formula for the inradius.
\(\sin\theta\)	sine theta	One of the trig ratios. In a right triangle, it is the height as a fraction of the hypotenuse, and it is used to get the vertical part from an angle. \(\sin 30^\circ = 0.5\) and \(\sin 90^\circ = 1\).
\(\cos\theta\)	cosine theta	One of the trig ratios. In a right triangle, it is the base as a fraction of the hypotenuse. \(\cos 60^\circ = 0.5\) and \(\cos 90^\circ = 0\). The "cosines" in the law of cosines is this function.
\(r\)	r (lowercase)	The radius of the incircle (the circle that touches all three sides from the inside).
\(R\)	R (uppercase)	The radius of the circumcircle (the circle through all three vertices).

Terms

opposite side	The side across from an angle is called the opposite side of that angle. On this page, side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C.
opposite angle	The angle across from a side is called the opposite angle of that side. On this page, angle A is opposite side a, angle B is opposite side b, and angle C is opposite side c.
angle sum	The fact that the three angles of a triangle always add up to 180° (the triangle angle sum theorem). If you know two angles, the third is found by subtraction.
triangle inequality	The condition for three lengths to form a triangle - the sum of any two sides must be greater than the third side. For example, 1, 2 and 4 do not form a triangle because 1 + 2 = 3 is less than 4.
law of cosines	The rule \(c^2 = a^2 + b^2 - 2ab\cos C\) for finding the third side from two sides and the included angle. It extends the Pythagorean theorem to triangles without a right angle, and rearranged, it gives an angle from three sides. It is taught in high school (Geometry or Precalculus).
law of sines	The rule that "side ÷ sine of its opposite angle" is the same for all three pairs and equals the diameter \(2R\) of the circumcircle. Once you know one side and its opposite angle, it bridges to the other sides and angles. It is taught in high school (Geometry or Precalculus).
Heron's formula	A formula that gives the area from the three side lengths alone, without any angle. It is named after Heron of Alexandria, an ancient Greek mathematician.
semiperimeter	Half the perimeter of a triangle. It appears in Heron's formula and in the formula for the inradius.
incircle	The inscribed circle, which touches all three sides of the triangle from the inside. Its center (the incenter) is where the three angle bisectors meet.
circumcircle	The circumscribed circle, which passes through all three vertices of the triangle. Its center (the circumcenter) is where the perpendicular bisectors of the three sides meet.
median	The segment from a vertex of a triangle to the midpoint of the opposite side. The three medians meet at one point (the centroid).
acute triangle	A triangle whose largest angle is less than 90°. By their angles, triangles are classified as acute, right or obtuse.
right triangle	A triangle whose largest angle is exactly 90°. By their angles, triangles are classified as acute, right or obtuse.
obtuse triangle	A triangle whose largest angle is greater than 90°. By their angles, triangles are classified as acute, right or obtuse.

Good to know before you start

Here is what helps you use the calculation on this page with real understanding, not just by pressing the button.
The angle sum and Heron's formula need only middle school math. The law of cosines and the law of sines are high school topics (Geometry or Precalculus).

Angles of a triangle (Grades 4–8)	Knowing that the angles of a triangle add up to 180° Knowing the properties of isosceles and equilateral triangles (angles opposite equal sides are equal)
Square roots (Grade 8)	Knowing that a square root is the number that, squared, gives back the original number, as in \(\sqrt{36} = 6\) Being able to estimate a value such as \(\sqrt{39}\), for example as between 6 and 7
The Pythagorean theorem (Grade 8)	Knowing that in a right triangle, \(c^2 = a^2 + b^2\) (the square of the hypotenuse equals the sum of the squares of the other two sides) Seeing the law of cosines as this theorem extended to triangles without a right angle
Right triangle trigonometry (Geometry)	Knowing that sine and cosine are ratios of the sides of a right triangle Being able to use the values for common angles, such as \(\sin 30^\circ = 0.5\), \(\cos 60^\circ = 0.5\) and \(\cos 90^\circ = 0\)
The law of sines and the law of cosines (Geometry or Precalculus)	Being able to choose the rule to use from which three values you know (SSS, SAS, ASA or AAS, and so on) Knowing that with two sides and an angle opposite one of them (SSA, the ambiguous case), there may be two possible triangles

How to calculate it in Excel

Copy the whole table below and paste it into cell A1 in Excel. It works as is.

Table to find the third angle (angle sum)

Angle A (degrees)	50
Angle B (degrees)	60
Third angle C (degrees)	=180-B1-B2

Table to find the third side with the law of cosines

Side a	5
Side b	7
Included angle C (degrees)	60
Third side c	=SQRT(B1^2+B2^2-2B1B2*COS(RADIANS(B3)))

Table to find another side with the law of sines

Side a	6
Its opposite angle A (degrees)	45
Angle B (degrees)	60
Side b	=B1*SIN(RADIANS(B3))/SIN(RADIANS(B2))

Table to find the area (two sides and the included angle)

Side a	8
Side b	5
Included angle C (degrees)	30
Area S	=B1B2SIN(RADIANS(B3))/2

Table to find the area (Heron's formula)

Side a	3
Side b	4
Side c	5
Semiperimeter s	=(B1+B2+B3)/2
Area S	=SQRT(B4(B4-B1)(B4-B2)*(B4-B3))

Table to find the inradius

Area S	6
Semiperimeter s	6
Inradius r	=B1/B2

Table to find the circumradius

Side a	3
Side b	4
Side c	5
Area S	6
Circumradius R	=B1B2B3/(4*B4)

After pasting, the upper rows are your inputs and the formula in the last row is calculated automatically.
Excel's COS and SIN take angles in radians, so the RADIANS function is placed in between to convert degrees to radians. SQRT is the square root.
For example, in the second table, B4 shows about 6.245. Just replace the input numbers with your own values.

How to calculate it in Google Sheets

Copy the whole table below and paste it into cell A1 in Google Sheets. It works as is.

Table to find the third angle (angle sum)

Angle A (degrees)	50
Angle B (degrees)	60
Third angle C (degrees)	=180-B1-B2

Table to find the third side with the law of cosines

Side a	5
Side b	7
Included angle C (degrees)	60
Third side c	=SQRT(B1^2+B2^2-2B1B2*COS(RADIANS(B3)))

Table to find another side with the law of sines

Side a	6
Its opposite angle A (degrees)	45
Angle B (degrees)	60
Side b	=B1*SIN(RADIANS(B3))/SIN(RADIANS(B2))

Table to find the area (two sides and the included angle)

Side a	8
Side b	5
Included angle C (degrees)	30
Area S	=B1B2SIN(RADIANS(B3))/2

Table to find the area (Heron's formula)

Side a	3
Side b	4
Side c	5
Semiperimeter s	=(B1+B2+B3)/2
Area S	=SQRT(B4(B4-B1)(B4-B2)*(B4-B3))

Table to find the inradius

Area S	6
Semiperimeter s	6
Inradius r	=B1/B2

Table to find the circumradius

Side a	3
Side b	4
Side c	5
Area S	6
Circumradius R	=B1B2B3/(4*B4)

The same formulas as in Excel work as is (Google Sheets has RADIANS, SIN, COS and SQRT under the same names).
Copy the whole table, paste it into cell A1, and replace the input numbers with your own values.

How to calculate it in Python

import math

side_a = 3.0   # side a
side_b = 4.0   # side b
side_c = 5.0   # side c (example where all three sides are known)

# Find the three angles with the law of cosines solved for the angle (in degrees)
angle_a = math.degrees(math.acos((side_b**2 + side_c**2 - side_a**2) / (2 * side_b * side_c)))
angle_b = math.degrees(math.acos((side_a**2 + side_c**2 - side_b**2) / (2 * side_a * side_c)))
angle_c = 180 - angle_a - angle_b

# Area with Heron's formula
half_perimeter = (side_a + side_b + side_c) / 2
area = math.sqrt(half_perimeter * (half_perimeter - side_a)
                 * (half_perimeter - side_b) * (half_perimeter - side_c))

inradius = area / half_perimeter                            # radius of the inscribed circle
circumradius = side_a * side_b * side_c / (4 * area)        # radius of the circumscribed circle

print(f"Angle A = {angle_a}°, angle B = {angle_b}°, angle C = {angle_c}°")
print(f"Area = {area}")
print(f"Inradius = {inradius}, circumradius = {circumradius}")

Runs with just math from the standard library. math.acos is the arccosine (cosine worked backward), and math.degrees converts radians to degrees. Replace the three sides at the top with your own values and run it.

How to write it in LaTeX and other math languages (copy and paste)

Angle sum of a triangle

A + B + C = 180°

A + B + C = 180^\circ

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>A</mi><mo>+</mo><mi>B</mi><mo>+</mo><mi>C</mi>
    <mo>=</mo>
    <mn>180</mn><mo>&#xB0;</mo>
  </mrow>
</math>

A + B + C = 180^@

angleA + angleB + angleC == 180

A + B + C = 180;

C = 180 - A - B;

A + B + C = 180°

Law of cosines (third side from two sides and the included angle)

c² = a² + b² − 2ab·cos C

c^{2} = a^{2} + b^{2} - 2ab\cos C

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <msup><mi>c</mi><mn>2</mn></msup>
    <mo>=</mo>
    <msup><mi>a</mi><mn>2</mn></msup>
    <mo>+</mo>
    <msup><mi>b</mi><mn>2</mn></msup>
    <mo>&#x2212;</mo>
    <mn>2</mn><mi>a</mi><mi>b</mi>
    <mi>cos</mi><mo>&#x2061;</mo><mi>C</mi>
  </mrow>
</math>

c^2 = a^2 + b^2 - 2ab cos C

cSide = Sqrt[a^2 + b^2 - 2 a b Cos[angleC Degree]]

c := sqrt(a^2 + b^2 - 2*a*b*cos(C*Pi/180));

c = sqrt(a^2 + b^2 - 2*a*b*cosd(C));

c^2 = a^2 + b^2 - 2ab cos(C)

Law of sines (the bridge between sides and opposite angles)

a/sin A = b/sin B = c/sin C = 2R

\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mfrac><mi>a</mi><mrow><mi>sin</mi><mo>&#x2061;</mo><mi>A</mi></mrow></mfrac>
    <mo>=</mo>
    <mfrac><mi>b</mi><mrow><mi>sin</mi><mo>&#x2061;</mo><mi>B</mi></mrow></mfrac>
    <mo>=</mo>
    <mfrac><mi>c</mi><mrow><mi>sin</mi><mo>&#x2061;</mo><mi>C</mi></mrow></mfrac>
    <mo>=</mo>
    <mn>2</mn><mi>R</mi>
  </mrow>
</math>

a/sin A = b/sin B = c/sin C = 2R

bSide = a Sin[angleB Degree]/Sin[angleA Degree]

b := a*sin(B*Pi/180)/sin(A*Pi/180);

b = a*sind(B)/sind(A);

a/sin(A) = b/sin(B) = c/sin(C) = 2R

Area (two sides and the included angle)

S = (1/2)·ab·sin C

S = \frac{1}{2}ab\sin C

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>S</mi>
    <mo>=</mo>
    <mfrac><mn>1</mn><mn>2</mn></mfrac>
    <mi>a</mi><mi>b</mi>
    <mi>sin</mi><mo>&#x2061;</mo><mi>C</mi>
  </mrow>
</math>

S = 1/2 ab sin C

area = (1/2) a b Sin[angleC Degree]

S := (1/2)*a*b*sin(C*Pi/180);

S = a*b*sind(C)/2;

S = (1/2)ab sin(C)

Heron's formula (area from three sides)

s = (a+b+c)/2,  S = √(s(s−a)(s−b)(s−c))

s = \frac{a+b+c}{2}, \quad S = \sqrt{s(s-a)(s-b)(s-c)}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>S</mi>
    <mo>=</mo>
    <msqrt>
      <mi>s</mi>
      <mo>(</mo><mi>s</mi><mo>&#x2212;</mo><mi>a</mi><mo>)</mo>
      <mo>(</mo><mi>s</mi><mo>&#x2212;</mo><mi>b</mi><mo>)</mo>
      <mo>(</mo><mi>s</mi><mo>&#x2212;</mo><mi>c</mi><mo>)</mo>
    </msqrt>
  </mrow>
</math>

S = sqrt(s(s-a)(s-b)(s-c))

area = Sqrt[s (s - a) (s - b) (s - c)] /. s -> (a + b + c)/2

s := (a+b+c)/2; S := sqrt(s*(s-a)*(s-b)*(s-c));

s = (a+b+c)/2; S = sqrt(s*(s-a)*(s-b)*(s-c));

S = √(s(s-a)(s-b)(s-c))

Radius of the inscribed circle (inradius)

r = S/s

r = \dfrac{S}{s}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>r</mi>
    <mo>=</mo>
    <mfrac><mi>S</mi><mi>s</mi></mfrac>
  </mrow>
</math>

r = S/s

inradius = area/s

r := S/s;

r = S/s;

r = S/s

Radius of the circumscribed circle (circumradius)

R = abc/(4S)

R = \dfrac{abc}{4S}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>R</mi>
    <mo>=</mo>
    <mfrac>
      <mrow><mi>a</mi><mi>b</mi><mi>c</mi></mrow>
      <mrow><mn>4</mn><mi>S</mi></mrow>
    </mfrac>
  </mrow>
</math>

R = (abc)/(4S)

circumradius = a b c/(4 area)

R := a*b*c/(4*S);

R = a*b*c/(4*S);

R = abc/(4S)

How to have ChatGPT do the calculation

You are a calculation assistant for triangles. Do the following calculation by actually running Python code, and base your answer only on the numbers from the execution result (do not answer by mental math or guessing).

In a triangle, side a = 7, side b = 10 and angle A = 40° (angle A is the angle opposite side a).
1. Use the law of sines to find angle B. Note that with these conditions there may be two possible triangles (the ambiguous case). If there are two solutions, show both.
2. For each solution, find the remaining angle C, side c and the area.

Show the formulas you used and the numbers from the execution result in a table.

How to Use

1

Enter your numbers

Type the numbers you want to calculate with into the input fields
2

Calculate

Press the "Calculate" button
3

Check the result

The result appears on the spot. The same page also explains the idea behind the calculation and the formula