Angle Sum Property of a Triangle: Definition, Properties, Proofs

Angle Sum Property of a Triangle: A triangle is one of the most commonly used shapes in geometry. A triangle is made up of three sides and three angles. The triangle’s elements are its sides and angles. All polygons have two kinds of angles: internal angles and outer angles. The triangle has three inner angles and six outside angles since it is the smallest polygon. ABC denotes a triangle with the vertices A, B, and C. There are many distinct types of triangles with varied angles and edges, but they always obey the triangle sum principles. The angle sum property of a triangle and the exterior angle property of a triangle are the two most essential properties.

The Angle Sum Property of a Triangle states that the sum of a triangle’s internal angles is 180 degrees. Interior angles are created at the vertex of a triangle where any two of its edges meet. The internal angle of a triangle is the angle formed by two sides of a triangle. It is also known as a triangle’s internal angle property. This property asserts that the sum of a triangle’s internal angles is 180°. The angle sum property formula for an ABC triangle is A+B+C = 180°. On this page, let us discuss everything about the Angle Sum Property of a Triangle. Read further to find more.

Angle Sum Property of Polygon

A polygon is a closed figure formed by straight line segments. The angles inside the polygon are known as interior angles. On the other hand, the angles outside the polygon are known as exterior angles formed by an extension of a side and its adjacent side. A polygon can have n number of sides.
We can get the sum of all the interior angles of it by using a specified formula.

The sum of the interior angles \( = (n – 2) \times {180^{\rm{o}}},\) where \(n\) is the number of sides.

The sum of the exterior angle of any polygon is \({360^{\rm{o}}}\).

We know that a triangle is a polygon with three sides.
Thus, using the above formula, we have \( = (3 – 2) \times {180^{\rm{o}}} = {180^{\rm{o}}}\) when \(n = 3\).

Angle Sum Property of Triangle

A triangle is the smallest polygon formed by three line segments, making the interior and exterior angles. An interior angle is an angle formed between two adjacent sides of a triangle. In contrast, an exterior angle is an angle formed between a side of the triangle and an adjacent side extending outward. There are different types of triangles, but for each type, the sum of the interior angles is \({180^{\rm{o}}}\). According to the angle sum property of a triangle, the sum of all three interior angles of a triangle is  \({180^{\rm{o}}},\) and the exterior angle of a triangle measures the same as the sum of its two opposite interior angles. Thus, the angle sum property of a triangle is useful for finding the measure of an unknown angle when the values of the other two angles are known.

Proof for the Sum of the Interior Angles of a Triangle

It is easier to prove the angle sum property using few geometrical concepts.

To prove: Sum of the interior angles of a triangle is \({180^{\rm{o}}}\)

In \(\Delta ABC\) given above, a line is drawn parallel to the side \(BC\) of \(\Delta ABC.\)
This line passes through vertex \(A\). Label this line as \(PQ\).
Since the straight angle measures \({180^{\rm{o}}}\),
Hence, \(\angle PAQ = {180^{\rm{o}}}\).
That is, \(\angle PAB + \angle BAC + \angle CAQ = {180^{\rm{o}}}\)
Let us mark this as equation \(1\).
\(\angle PAB + \angle BAC + \angle CAQ = {180^{\rm{o}}}………\left( 1 \right)\)
Now, we need to prove \(∠PAB=∠ABC\) and \(∠CAQ=∠ACB\)
As \(PQ || BC\) and \(AB\) is a transversal, then alternate interior angles are equal/congruent
\(\therefore \angle PAB = \angle ABC\)
Let us mark this as equation \(2\).
\(∠PAB=∠ABC…..(2)\)
Similarly, as \(PQ || BC\) and \(AC\) is a transversal, then alternate interior angles are equal/congruent. 
\(∴ ∠CAQ=∠ACB\)
Let us mark this as equation \(3\).
\(∠CAQ=∠ACB…..(3)\)
So, the theorem used here is that the alternate interior angles are equal/congruent if a transversal intersects the lines.
Now, using equations \(2\) and \(3\) marked above, substitute \(∠ABC\) for \(∠PAB\) and \(∠ACB\) for \(∠CAQ\)  in equation \(1\): 
So, the equation \(∠PAB + ∠BAC + ∠CAQ = {180^{\rm{o}}}\)
becomes
\(∠ABC + ∠BAC + ∠ACB = {180^{\rm{o}}}\)
Let this be marked as equation \(4.\)
\(∠ABC + ∠BAC + ∠ACB = {180^{\rm{o}}}………..\left( 4 \right)\)
Hence, if we consider \(\Delta ABC,\) equation \(4\) implies that the sum of the interior angles of  \(\Delta ABC\) is \({180^{\rm{o}}}\). We can also write this as
\(∠A + ∠B + ∠C = {180^{\rm{o}}}.\)
Thus, it is proved that the sum of all the interior angles of a triangle is \({180^{\rm{o}}}.\)

Proof for the Relation Between Exterior Angle and the Sum of the Opposite Interior Angles of a Triangle

\(∠ACB\) and \(∠ACD\) form a linear pair of angles since they represent the adjacent angles on a straight line.
Thus, \(\angle ACB + \angle ACD = {180^{\rm{o}}}………\left( 1 \right)\) (linear pair axiom)
Also, from the angle sum property, it follows that:
\(\angle ACB + \angle BAC + \angle CBA = {180^{\rm{o}}}………\left( 2 \right)\) (angle sum property of triangle)
From equation \((1)\) and \((2),\) it follows that:
\(∠ACB+∠ACD=∠ACB+∠BAC+∠CBA\)
Now, cancelling \(∠ACB\) from both the sides we have,
\(∠ACD=∠BAC+∠CBA\)

This property can also be proved using the concept of parallel lines as follows:

In \(ABC,\) side \(BC\) is extended.
A line \(CE\) is drawn parallel to the side \(AB.\)
Since \(BA || CE\) and \(AC\) is the transversal,
\(∠CAB=∠ACE………(3)\) (Pair of alternate angles)
Also, \(BA || CE\) and \(BD\) is the transversal
Therefore, \(∠ABC=∠ECD……….(4)\) (Corresponding angles)
We have, \(\angle ACB + \angle BAC + \angle CBA = {180^{\rm{o}}}………\left( 5 \right)\)
Since the sum of angles on a straight line is \({180^{\rm{o}}}\)
Therefore, \(\angle ACB + \angle ACE + \angle ECD = {180^{\rm{o}}}………\left( 6 \right)\)
Since, \(∠ACE+∠ECD=∠ACD\)
Substituting this value in equation \((6)\)
\(\angle ACB + \angle ACD = {180^{\rm{o}}}………..\left( 7 \right)\)
From the equations \((5)\) and \((7)\) we get,
\(∠ACB+∠ACD=∠ACB+∠BAC+∠CBA\)
Now, cancelling \(∠ACB\) from both the sides we have,
\(∠ACD=∠BAC+∠CBA\)
Hence, it can be observed that the exterior angle of a triangle equals the sum of its opposite interior angles.

Application of Angle Sum Property of Triangle

We can use the angle sum property of the triangle to find the sum of the interior angles of another polygon. Since every polygon can be divided into triangles, the angle sum property can be extended to find the sum of the angles of all polygons. Let us see how this is applicable in quadrilaterals.

Angle Sum Property of a Quadrilateral

A diagonal of a quadrilateral divides a quadrilateral into two triangles. So, the sum of angles of a quadrilateral will be equal to the sum of angles of two triangles.
That is, the sum of the interior angles of a quadrilateral is \({360^{\rm{o}}}\).
Let’s prove that the sum of all the four angles of a quadrilateral is \({360^{\rm{o}}}\).

We know that the sum of angles in a triangle is \({180^{\rm{o}}}\) from the first proof
Now, consider \(△ADC,\)
\(\angle ADC + ∠DAC + ∠DCA = {180^{\rm{o}}}………..\left( 1 \right)\) (Sum of the interior angles of a triangle)
Now, consider triangle \(△ABC,\)
\(\angle ABC + ∠BAC + ∠BCA = {180^{\rm{o}}}………..\left( 2 \right)\) (Sum of the interior angles of a triangle)
On adding both equations \((1)\) and \((2),\) we have,
\((\angle ADC + \angle DAC + \angle DCA) + (\angle ABC + \angle BAC + \angle BCA) = {180^{\rm{o}}} + {180^{\rm{o}}}\)
\( \Rightarrow \angle ADC + (\angle DAC + \angle BAC) + (\angle BCA + \angle DCA) + \angle ABC = {360^{\rm{o}}}\)
We see that \((∠DAC+∠BAC)=∠DAB\) and \((∠BCA+∠DCA)=∠BCD.\)
Substituting them we have,
\(\angle ADC + ∠DAB + ∠BCD + \angle ABC = {360^{\rm{o}}}\)
Hence, the sum of angles of a quadrilateral is \({360^{\rm{o}}}\) which is known as the angle sum property of quadrilaterals.

Solved Examples – Angle Sum Property of a Triangle

Q.1. If the sum of two interior angles is \({110^{\rm{o}}}\), find the third angle.
Ans: Given, the sum of two interior angles is \({110^{\rm{o}}}\).
Let us assume the third angle is \(x\).
We know that sum of three interior angles is \({180^{\rm{o}}}\).
Thus, \(x + {110^{\rm{o}}} = {180^{\rm{o}}} \Rightarrow x = {180^{\rm{o}}} – {110^{\rm{o}}} = {70^{\rm{o}}}\)

Q.2. If the angles of a triangle are in the ratio \(3:4:5,\) determine the value of the three angles.
Ans: Let the angles be \(3x,\, 4x\) and \(5x\).
According to the angle sum property of the triangle,
\(3x + 4x + 5x = {180^{\rm{o}}},\)
\( \Rightarrow 12x = {180^{\rm{o}}},\)
\( \Rightarrow x = {15^{\rm{o}}}\)
Thus, the three angles will be \(3x = 3 \times {15^{\rm{o}}} = {45^{\rm{o}}},4x = 4 \times {15^{\rm{o}}} = {60^{\rm{o}}},5x = 5 \times 15 = {75^{\rm{o}}}\).
Therefore, the three angles are \({45^{\rm{o}}},{60^{\rm{o}}},{75^{\rm{o}}}\)

Q.3. In an isosceles \(\Delta DEF,\), if \(∠D = {120^{\rm{o}}},\) what is the measurement of \(∠F\)?
Ans: Given, \(∠D = {120^{\rm{o}}},\)
Two angles can not be \({120^{\rm{o}}}\) as the sum of all interior angles is \({180^{\rm{o}}}\).
So, \(∠F\) can not be \({120^{\rm{o}}}\).
We can say, \(\angle F + \angle E = {180^{\rm{o}}} – {120^{\rm{o}}} = {60^{\rm{o}}}\)
\(∠F=∠E\) (the triangle is isosceles)
Therefore, \(\angle F = \frac{{{{60}^{\rm{o}}}}}{2} = {30^{\rm{o}}}\)

Q.4. If an exterior angle is \({100^{\rm{o}}}\) and one of its opposite interior angles is \({60^{\rm{o}}}\), find the other two angles.
Ans: Given the exterior angle is \({100^{\rm{o}}}\).
Let us say, one opposite interior angle to the exterior angle is \(x\).
So, \(x + {60^{\rm{o}}} = {100^{\rm{o}}}\) (an exterior angle is equal to the sum of its opposite interior angles)
\( \Rightarrow x = {40^{\rm{o}}}\)
Therefore, the other angle \( = {180^{\rm{o}}} – \left( {{{60}^{\rm{o}}} + {{40}^{\rm{o}}}} \right) = {80^{\rm{o}}}\)
Hence, the other two angles of the triangle are \({80^{\rm{o}}},{40^{\rm{o}}}.\)

Q.5. One of the acute angles of a right triangle is \({48^{\rm{o}}}\). Find the measurement of the other acute angle.
Ans: Given, one of the acute angles is \({48^{\rm{o}}}\).
The other angle of the triangle is \({90^ \circ }.\)
Let us say the other acute angle is \(x\).
So, \(x + {90^{\rm{o}}} + {48^{\rm{o}}} = {180^{\rm{o}}} \Rightarrow x = {180^{\rm{o}}} – {138^{\rm{o}}} = {42^{\rm{o}}}\)
Hence, the other acute angle is \({42^{\rm{o}}}\)

Summary

We have learned in this article that the sum of the interior angles of a triangle is equal to \({180^{\rm{o}}},\) and an exterior angle of a triangle is equal to the sum of two opposite interior angles.
These two properties are applicable for every type of triangle.

Frequently Asked Questions (FAQs) on Angle Sum Property of a Triangle

Q.1. Explain the angle sum property of a triangle.
Ans: The angle sum property of a triangle states that the sum of all three interior angles of a triangle is 180°, and the exterior angle of a triangle measures the same as the sum of its two opposite interior angles.

Q.2. What is the formula of the sum of the interior angles of a polygon with \(n\) numbers of sides?
Ans: The formula of the sum of the interior angle \( = (n – 2) \times {180^{\rm{o}}}\)

Q.3. What is the exterior angle property of a triangle?
Ans: The exterior angle property says that if we extend one of the sides of a triangle, we will get an exterior angle that is equal to the sum of the opposite interior angles.

Q.4. How to implement/prove the angle sum property of a triangle?
Ans: To implement/prove the angle sum property, we need to construct a line that is parallel to the base of the triangle. Then using the properties of parallel lines and linear pair axiom, we can implement/prove it.

Q.5. What are the applications of the angle sum property of a triangle?
Ans: If we can divide a polygon into triangular parts then, we can use this concept of the angle sum property of a triangle to find the sum of the interior angles of a polygon. For example, we can prove that the sum of all the interior angles of a quadrilateral is \({360^{\rm{o}}}\).