Solve the Triangle Given Its Three Sides' Lengths: 73, 78 and 80. Triangle Calculator: Angles, Area, (Semi)Perimeter, Heights, Medians, Circles Radii

• The given triangle is solved below step by step, with full explanations of the used formulas, clear explanations on the used terms and with illustrative pictures.
• At the end, all the calculated data of the triangle is summarized in one section.
• The order of steps of solving the triangle: 1) calculating the perimeter, 2) the semiperimeter, 3) the measures of the angles, 4) the area - calculated by two methods, 5) the lengths of the three heights of the triangle, 6) the lengths of the three medians of the triangle, 7) the radius of the inscribed circle of the triangle and 8) the radius of the circumscribed circle.

Scalene triangle

• This is a scalene triangle, with all the lengths of its sides different from each other, without being a right triangle.
• The triangle is not equilateral, nor isosceles, nor right-angled, things that can be determined easily from the lengths of the sides.

These formulas are directly derived from, and a rearrangement for angles, of the Law of Cosine (or generalized Pythagorean Theorem).

• m(∡A) = arccos(^{(b2 + c2 - a2)} / _{(2 × b × c)})
• m(∡B) = arccos(^{(a2 + c2 - b2)} / _{(2 × a × c)})
• m(∡C) = arccos(^{(a2 + b2 - c2)} / _{(2 × a × b)})
• Where:
• a = the side of the triangle opposite angle ∠A
• b = the side of the triangle opposite angle ∠B
• c = the side of the triangle opposite angle ∠C

• Why this formula? Proof of the Law of Cosine

• In any triangle △ABC, which is not right-angled, with side lengths BC = a, AC = b and AB = c, we draw from the vertex A the perpendicular to the side BC, which intersects BC in point D. We note the lengths of the sides AD = h_a, BD = a1 and CD = a2, where a1 + a2 = a.
• 

  Proof of the Law of Cosine

• In the right triangle △ABD we have the following trigonometric relations:
• sin(∠B) = ^h_a/_c => h_a = c × sin(∠B)
• cos(∠B) = ^a1/_c => a1 = c × cos(∠B)
• In the right triangle △ADC we apply the Pythagorean Theorem:
• b² = h_a² + a2² <=>
• b² = h_a² + (a - a1)²

• ---

• Substitute for h_a and a1 in the formula above:
• b² = c² × sin²(∠B) + (a - c × cos(∠B))² <=>
• b² = c² × sin²(∠B) + a² - 2 × a × c × cos(∠B) + c² × cos²(∠B) <=>
• b² = c² × (sin²(∠B) + cos²(∠B)) + a² - 2 × a × c × cos(∠B) <=>
• b² = c² × (^h_a2/_c² + ^a12/_c²) + a² - 2 × a × c × cos(∠B) <=>
• b² = c² × ^{(h_a2 + a12)}/_c² + a² - 2 × a × c × cos(∠B) <=>

• ---

• In the right triangle △ABD h_a² + a1² = c² (Pythagora's Theorem) =>
• ^{(h_a2 + a12)}/_c² = 1 =>

• ---

• b² = c² + a² - 2 × a × c × cos(∠B) <=>
• cos(∠B) = ^{(a2 + c2 - b2)} / _{(2 × a × c)}
• Apply function arccos to both sides of the equality:
• arccos(cos(∠B)) = arccos(^{(a2 + c2 - b2)} / _{(2 × a × c)}) =>
• m(∡B) = arccos(^{(a2 + c2 - b2)} / _{(2 × a × c)})
• The same process is used to calculate the measures of angles ∠A and ∠C.

• Last three definitions: π. Radians. Degrees

• Definition: π represents the constant ratio of a circle's circumference (circle's length), C, to its diameter, d (d = 2 × r, where r is the radius of the circle), which is approximately 3.14159, commonly estimated as the fraction ²²/₇.
  π = ^C/_d = ^C/_{(2 × r)}
• 

  Angle measurement: π. Radians. Degrees

• Definition: A radian is a standard unit of angle measurement. A radian is equal to the angle formed when the length of the arc of a circle is equal to the radius of the same circle. There are exactly 2π radians in the circumference (the length) of a circle (any circle). 1 radian is roughly 57.296° (degrees). An angle of 360° = 2π radians, 180° = π radians, 90° = ^π/₂ radians, 45° = ^π/₄ radians.
• Definition: Angles are usually measured in degrees (°). A degree is a standard unit of angle measurement. The angle formed by a complete circle is 360°. An obtuse angle is between (90° and 180°). A right angle is 90°. An acute angle is between (0° and 90°).

• ∡A = 55.018122807495°
• ∡B = 61.098533258644°
• ∡C = 63.883343933865°

Note: The sum of the angles is approximately 180° instead of exactly 180°. This is because we performed calculations that required rounding: calculating the arccos and converting from radians to degrees.

• This method requires more steps to get to the result than the second one.
• In theory, both calculation formulas will arrive at exactly the same result. But in practice, when using calculation tools that involve rounding and approximations, it is better to use the calculation method that involves fewer steps, and fewer approximations.
• This method is presented here for demonstrative purposes, we will calculate the area using the second method, which involves fewer steps and fewer approximations.

• The classic formula:

• Area = ¹/₂ × a × h_a
• Where:
• a = the side of the triangle opposite angle ∠A
• h_a = the height AD from the vertex A to the side "a" (BC), where D belongs to the line containing the segment BC.
• 

  Height h_a expressed as a function of the side "b" and the sine of angle C

• == But we don't know the height h_a. What to do next? ==
• Using trigonometry, in the right triangle △ADC (∡D = 90°), formed to the right of the height h_a:
• sin(∠C) = ^h_a / _b
• It follows that h_a = b × sin(∠C).
• Substitute the height h_a into the classic formula of the area and get:

• Trigonometric formula:

• Area = ¹/₂ × a × b × sin(∠C)

Note: Area will be calculated using the 2^nd method, which involves fewer steps and approximations.

• Heron's Formula

• Given the lengths of its three sides, the area of a triangle can be calculated using the Heron's Formula:
• Area² = s × (s - a) × (s - b) × (s - c)
• Where:
• a, b and c are the lengths of the three sides of the triangle
• s is the semiperimeter of the triangle, already calculated above

• Step-by-Step Pythagorean Proof of Heron's Formula

• Consider a triangle △ABC with its sides BC = a, AC = b and AB = c. Draw the height from vertex A, opposite to side "a", AD ⊥ BC, D ∈ BC, AD = h. AD splits the base "a" into two segments, BD = x and CD = a - x.
• 

  Triangle △ABC for Heron's Formula

• Apply the Pythagorean Theorem to the right-angled triangle △ADB:
• h² + x² = c² =>
• h² = c² - x²
• Apply the Pythagorean Theorem to the right-angled triangle △ADC:
• h² + (a - x)² = b² =>
• h² = b² - (a - x)²
• Equate the two expressions for h² and solve for the segment "x":
• c² - x² = b² - (a - x)² <=>
• c² - x² = b² - (a² - 2 × a × x + x²) <=>
• c² - x² = b² - a² + 2 × a × x - x² <=>
• c² = b² - a² + 2 × a × x <=>
• 2 × a × x = a² + c² - b² <=>
• x = ^{(a2 + c2 - b2)}/_{(2 × a)}
• Substitute "x" back into the formula of h² and calculate the height "h":
• h² = c² - x² <=>
• h² = c² - (^{(a2 + c2 - b2)}/_{(2 × a)})²
• Factor the expression using the difference of squares rule, c² - x² = (c - x) × (c + x):
• h² = ^{((2 × a × c)2 - (a2 + c2 - b2)2)} / _{(2 × a)})² <=>
• h² = ^{(2 × a × c - a2 - c2 + b2) × (2 × a × c + a2 + c2 - b2)} / _{(2 × a)²} <=>
• h² = ^{(b2 - (a2 - 2 × a × c + c2)) × ((a2 + 2 × a × c + c2) - b2)} / _{(2 × a)²} <=>
• h² = ^{(b2 - (a - c)2) × ((a + c)2 - b2)} / _{(2 × a)²} <=>
• h² = ^{(b - a + c) × (b + a - c) × (a + c - b) × (a + c + b)} / _{(2 × a)²} <=>
• h² = ^{(a + b + c) × (b + c - a) × (a + c - b) × (a + b - c)} / _{(2 × a)²}
• Substitute for h² into the Area Formula:
• Area = ¹/₂ × a × h => Area² = ¹/₄ × a² × h² =>
• Area² = ¹/₄ × a² × ^{(a + b + c) × (b + c - a) × (a + c - b) × (a + b - c)} / _{(4 × a²)} <=>
• [1] Area² = ¹/₁₆ × (a + b + c) × (b + c - a) × (a + c - b) × (a + b - c)
• Use the semi-perimeter s = ^{(a + b + c)}/₂ to express the factors of the formula [1] above:
• => a + b + c = 2 × s
• => b + c - a = 2 × s - 2 × a = 2 × (s - a)
• => a + c - b = 2 × s - 2 × b = 2 × (s - b)
• => a + b - c = 2 × s - 2 × c = 2 × (s - c)
• Rewrite formula [1]:
• Area² = ¹/_2⁴ × 2 × s × 2 × (s - a) × 2 × (s - b) × 2 × (s - c) <=>
• Area² = s × (s - a) × (s - b) × (s - c) - Heron's Formula

Area ≈ 2,556.320292432073 ≈ 2,556.32 square units

• Definition: The height of a triangle is a straight line segment drawn from a vertex, perpendicular (at a 90-degree angle) to the opposite side. Every triangle has three heights. All three heights always cross each other at one single point, which is called the orthocenter.
• 

  The three heights of the triangle

• In the triangle △ABC h_a is the height starting from the vertex A, perpendicular to the side BC, h_a ⊥ BC. h_b is the height starting from the vertex B, perpendicular to side AC, h_b ⊥ AC. h_c starts from the vertex C, perpendicular to the side AB, h_c ⊥ AB. They all intersect at point H, the orthocenter.

• ---

• Knowing the area and sides of the triangle, we can calculate the heights h_a, h_b and h_c from vertices A, B and C to the sides a, b and c respectively.
• Starting from the classic formula of area:
• Area = ¹/₂ × a × h_a =>
• h_a = ^{2 × Area} / _a ≈ 70.036172395399 ≈ 70.04
• Area = ¹/₂ × b × h_b =>
• h_b = ^{2 × Area} / _b ≈ 65.546674164925 ≈ 65.55
• Area = ¹/₂ × c × h_c =>
• h_c = ^{2 × Area} / _c ≈ 63.908007310802 ≈ 63.91

• Definition: A median of a triangle is a line segment connecting a vertex of the triangle to the midpoint of the opposite side. Every triangle has three medians that meet at a single center point, called centroid.
• 

  The three medians of the triangle

• In the triangle △ABC m_a is the median connecting the vertex A to the middle of side BC, m_b is the median connecting the vertex B to the middle of side AC and m_c is the median connecting the vertex C to the middle of the side AB. They all intersect at point G, the centroid.
• The centroid G divides the lengths of the three medians as follows:
• ^AG/_AM = ^BG/_BP = ^CG/_CN = ²/₃
• ^GM/_AM = ^GP/_BP = ^GN/_CN = ¹/₃
• ^GM/_AG = ^GP/_BG = ^GN/_CG = ¹/₂
  

• Apollonius' Theorem

• == Proof of the Apollonius' Theorem at the end of the section. ==
• In a triangle △ABC, for which we know the lengths of the sides "a", "b" and "c", having the opposite vertices A, B and C, respectively, there is a relationship between the lengths of the sides and the lengths of the medians m_a, m_b and m_c, where m_a is drawn from vertex A to side "a", m_b is drawn from vertex B to side "b", and m_c is drawn from vertex C to side "c":
• b² + c² = 2 × (m_a² + (^a/₂)²) =>
• m_a² = ^{(b2 + c2)}/₂ - (^a/₂)²= 4,909.75 =>
• Median m_a ≈ 70.069608247799 ≈ 70.07

• ---

• a² + c² = 2 × (m_b² + (^b/₂)²) =>
• m_b² = ^{(a2 + c2)}/₂ - (^b/₂)²= 4,343.5 =>
• Median m_b ≈ 65.905234996926 ≈ 65.91

• ---

• a² + b² = 2 × (m_c² + (^c/₂)²) =>
• m_c² = ^{(a2 + b2)}/₂ - (^c/₂)²= 4,106.5 =>
• Median m_c ≈ 64.081978745978 ≈ 64.08

• ---

• Proof of the Apollonius' Theorem

• We prove Apollonius' Theorem using the Law of Cosines, which we proved above, when we calculated the angles' measures.
• 

  The three medians of the triangle

• Let median AM, of lenght m_a, divide the side BC, of length "a", into two equal parts, of length BM = CM = ^a/₂. Also AB = c, AC = b, BC = a.
• Let the measure of the angle ∡AMB = θ. Since two angles on a straight line add up to 180° then the adjacent angle ∡AMC = 180° - θ.
• => cos(180° - θ) = - cos(θ)
• Apply the Law of Cosines to the triangle △ABM, making sure to also include the angle ∡AMB = θ in the formula:
• c² = m_a² + (^a/₂)² - 2 × m_a × ^a/₂ × cos(θ) =>
• [1] c² = m_a² + (^a/₂)² - m_a × a × cos(θ)
• Apply the Law of Cosines to the triangle △ACM, making sure to also include the angle ∡AMC = 180° - θ in the formula:
• b² = m_a² + (^a/₂)² - 2 × m_a × ^a/₂ × cos(180° - θ)
• cos(180° - θ) = - cos(θ)
• The above relation becomes:
• [2] b² = m_a² + (^a/₂)² + m_a × a × cos(θ)
• Add the two equations [1] and [2] together. The terms m_a × a × cos(θ) and - m_a × a × cos(θ) cancel each other:
• b² + c² = 2 × (m_a² + (^a/₂)²)

• Intro

• The bisector of an angle is a half-right inside the angle, with the origin at its apex, which forms two smaller congruent angles with the sides of the angle. Any point belonging to the bisector is equidistant from both sides of the angle.
• 

  Bisector of an angle

• In any triangle the bisectors of the angles cross at the exact same point, which is the center of the circle tangent to the sides of the triangle, called the circle inscribed in the triangle. This is the largest circle that fits inside the triangle.
• 

  Bisectors and the circle inscribed in the triangle

• In the triangle △ABC, AA' is the bisector of the angle ∠BAC => ∠BAA' = ∠CAA' = ^∠BAC / ₂
• BB' is the bisector of the angle ∠ABC => ∠ABB' = ∠CBB' = ^∠ABC / ₂
• CC' is the bisector of the angle ∠ACB => ∠ACC' = ∠BCC' = ^∠ACB / ₂

• ---

• Bisectors AA', BB' and CC' intersect at point O, which is the center of the circle inscribed in the triangle. Why is this happening?

• ---

• O is a point on the bisector AA' so it is equidistant from the line segments AB and AC => [1] the distance from the point O to the segment AB is OP, where OP ⊥ AB (⊥ is the symbol for perpendicular segments), [2] the distance from the point O to the segment AC is ON, where ON ⊥ AC and [3] OP ≅ ON.
• O is a point on the bisector BB' so it is equidistant from the line segments AB and BC => [1] the distance from the point O to the segment AB is OP, where OP ⊥ AB, [2] the distance from the point O to the segment BC is OM, where OM ⊥ BC and [3] OP ≅ OM.
• From the relations above, it follows that OM = ON = OP = r, where r is the radius of the circle inscribed in the triangle

• To calculate the radius of the circle inscribed in the triangle, divide the area of ​​the triangle by its semiperimeter:
• r = ^Area / _s
• Where:
• r = radius of the circle inscribed in the triangle
• s is the semiperimeter of the triangle, already calculated above

• ---

• Why this formula? How to prove it?
• The area of ​​the triangle △ABC = area △OAB + area △OBC + area △OAC = ¹/₂ × OP × AB + ¹/₂ × OM × BC + ¹/₂ × ON × AC = ¹/₂ × r × AB + ¹/₂ × r × BC + ¹/₂ × r × AC = ¹/₂ × r × (AB + BC + AC) = ¹/₂ × r × Perimeter = ^Perimeter/₂ × r = Semiperimeter × r =>
  r = ^Area / _{Semiperimeter} = ^Area / _s

• ---

• r = ^{2,556.320292432073} / _115.5 ≈ 22.13264322452 ≈ 22.13

• Intro

• The perpendicular bisector of a segment is the line perpendicular to the segment that passes through its middle. Any point belonging to the perpendicular bisector is equidistant from the ends of the segment.
• 

  The perpendicular bisectors and circumcircle of the triangle

• In any triangle the perpendicular bisectors of the sides are concurrent in a point which is the center of the circle passing through the vertices of the triangle, called the circumscribed circle of the triangle.
• In the triangle △ABC, OM1 is the perpendicular bisector of segment BC => BM1 = CM1 = ^BC / ₂ = ^a / ₂, where a is the length of segment BC.
• OM2 is the perpendicular bisector of segment AC => AM2 = CM2 = ^AC / ₂ = ^b / ₂, where b is the length of segment AC.
• OM3 is the perpendicular bisector of segment AB => AM3 = BM3 = ^AB / ₂ = ^c / ₂, where c is the length of segment AB.

• ---

• The perpendicular bisectors OM1, OM2 and OM3 intersect at point O, which is the center of the circumscribed circle of the triangle. Why is this happening?

• ---

• O is a point on the perpendicular bisector OM1 => BM1 = CM1 = ^a/₂, OM1 ⊥ BC (⊥ is the symbol for segments or perpendicular lines), and OM1 common segment => right triangles △OBM1 ≅ △OCM1 => OB ≅ OC.
• O is a point on the perpendicular bisector OM2 => AM2 = CM2 = ^b/₂, OM2 ⊥ AC, and OM2 common segment => right triangles △OAM2 ≅ △OCM2 => OA ≅ OC.
• From the relations above, it follows that OA = OB = OC = R, where R is the radius of the circle circumscribing the triangle.

• To calculate R, the radius of the circle circumscribing the triangle, apply the formula:
• R = ^{a × b × c} / _{(4 × Area)}
• Where:
• a, b and c are the lengths of the three sides of the triangle

• ---

• Why this formula? How to prove it?
• The Law of Sines establishes the relationship between the lengths of the sides a, b and c of a triangle, for example △ABC, and the sines of the opposite angles, ∠A, ∠B and ∠C:
• [1] ^a/_sin(∠A) = ^b/_sin(∠B) = ^c/_sin(∠C) = 2 × R
• R is the radius of the circumscribed circle of the triangle △ABC
• We also use the Trigonometric Formula of the Area which we proved above, when we have been calculating the area:
• [2] Area = ¹/₂ × b × c × sin(∠A)

• ---

• From relationship [1] we can deduce that:
• [3] sin(∠A) = ^a/_{(2 × R)}
• Substitute for sin(∠A) in the relationship [2]:
• Area = ¹/₂ × b × c × ^a/_{(2 × R)} = ^{a × b × c}/_{(4 × R)} =>
• R = ^{a × b × c}/_{(4 × Area)}

• ---

• R = ^{73 × 78 × 80}/_{(4 × 2,556.320292432073)} ≈ 44.548408248035 ≈ 44.55

• 

  Triangle △ABC with side lengths: BC (a) = 73, AC (b) = 78, AB (c) = 80

Symbols used: + addition, - subtraction, × multiplication, / division, = equal, ≈ approximately, => logical implication, ∠ angle, ∡ measured angle, m(∡A) angle's measure, △ triangle, ⌒ arc of a circle, ⊥ perpendicular, ≅ congruence, π the constant ratio of a circle's circumference to its diameter, ° unit of the measured value of an angle and it is called degree(s), ∈ it belongs to, x² x raised to the second power or x squared = x × x, sin(∠A) sine of angle A, cos(∠A) cosine of angle A, arccos(ratio) arccosine of given ratio.

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