Exterior Angle of a Triangle and its Properties: Examples

Exterior Angles in Triangles (GMAT/GRE/CAT/Bank PO/SSC CGL) | Don’t Memorise
Exterior Angles in Triangles (GMAT/GRE/CAT/Bank PO/SSC CGL) | Don’t Memorise

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August 21, 2023

The angles created between the side of the polygon and the extended adjacent side of the polygon are known as exterior angles. According to the external angle theorem, when a triangle’s side is stretched, the resulting exterior angle is equal to the sum of the measurements of the triangle’s two opposed internal angles. In geometry, this topic, Exterior Angle of a Triangle and its Properties, is important for students to learn.

In a triangle, the exterior angle theorem can compute the measure of an unknown angle. We must first determine the triangle’s exterior angle and then the two adjacent remote interior angles to use the theorem. This article will discuss the definition of an exterior angle and its properties and theorems based on the exterior angle. Continue reading to know more.

A linear pair of angles are formed when two lines intersect at a single point. The angles are considered linear if they are adjacent to each other following the intersection of the two lines. The angles are supplementary if they form a linear pair and their measures add up to \({180^ \circ }.\) As a result, a linear pair of angles always equals \({180^ \circ }.\)

If we consider the triangle \(\Delta ABC,\) the two opposite angles \(\angle ABC\) and \(\angle BAC\) are called interior opposite angles of \(\angle ACB.\)

When a line is stretched from a closed shape’s outer edge, it generates an exterior angle. It is produced by the shape’s side and the adjacent extended side. When one of the triangle’s sides is extended, an exterior angle is formed.

The angle formed by one side of a triangle and the extension of an adjacent side is called an exterior (or external) angle.

If the side \(BC\) of a triangle \(ABC\) is produced to form ray \(BD,\) then \(\angle ACD\) is called an exterior angle of \(\Delta ABC\) at \(C\) and is denoted by ext. \(\angle ACD.\)

Concerning ext. \(\angle ACD\) of \(ABC\) at \(C,\) the angles \(A\) and \(B\) are called remote interior angles or interior opposite angles.

If we produce side \(AC\) to form ray \(AE,\) then \(\angle BCE\) is also an exterior angle of \(\Delta ABC\) at \(C.\) These two angles viz. ext \(\angle ACD\) and ext. \(\angle BCE\) are vertically opposite angles.

Therefore, ext.\(\angle ACD = \) ext. \(\angle BCE\)

Also, angles \(A\) and \(B\) are the interior opposite angles concerning ext. \(\angle BCE.\)

It follows from the above discussion that at each vertex of a triangle, there are two exterior angles of the triangle, and these two angles are equal. An exterior angle of a triangle is closely related to the interior opposite angles, as proved in the below theorem.

Theorem 1: If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.

Given: A triangle \(\Delta ABC.D\) is the point on \(BC\) produced, forming exterior angle \(\angle 4.\)

To prove: \(\angle 4 = \angle 1 + \angle 2\) i.e., \(\angle ACD = \angle CAB + \angle CBA\)

Proof: In triangle \(\Delta ABC,\) we have

\(\angle 1 + \angle 2 + \angle 1 = {180^ \circ }…..\left({\text{i}} \right)\)

Also, \(\angle 3 + \angle 4 = {180^ \circ }\) (Since \(\angle 3\) and \(\angle 4\) forms a linear pair) \(……….\left({{\text{ii}}} \right)\)

From \(\left({\text{i}} \right)\) and \(\left({{\text{ii}}} \right),\) we have

\(\angle 1 + \angle 2 + \angle 3 = \angle 3 + \angle 4\)

\( \Rightarrow \angle 1 + \angle 2 = \angle 4\)

Hence, \(\angle 4 = \angle 1 + \angle 2\) i.e., \(\angle ACD = \angle CAB + \angle CBA\)

Corollary: An exterior angle of a triangle is greater than either of the interior opposite angles.

Proof: Let \(ABC\) be a triangle whose side \(BC\) is produced to form an exterior angle \(\angle 4.\) Then,
\(\angle 1 + \angle 2 = \angle 4\)
\( \Rightarrow \angle 4 > \angle 1\) and \(\angle 4 > \angle 2,\) i.e., \(\angle ACD > \angle CAB\) and \(\angle ACD > \angle CBA\)

Theorem 2 (Exterior Angle Inequality Theorem): The sides \(AB\) and \(AC\) of a \(\Delta ABC\) are produced to \(P\) and \(Q\) respectively. If the bisectors of \(\angle PBC\) and \(\angle QCB\) intersect at \(O,\) then \(\angle BOC = {90^ \circ } – \frac{1}{2}\angle A\)

Given: A \(\Delta ABC\) in which sides \(AB\) and \(AC\) are produced to \(P\) and \(Q\) respectively. The bisectors of \(\angle PBC\) and \(\angle QCB\) intersect at \(O.\)
To prove: \(\angle BOC = {90^ \circ } – \frac{1}{2}\angle A\)

Proof: Since \(\angle ABC\) and \(\angle CBP\) form a linear pair.
\(\therefore \angle ABC + \angle CBP = {180^ \circ }\)
\( \Rightarrow \angle B + 2\angle 1 = {180^ \circ }\) (\(\because BO\) is the bisector of \(\angle CBP.\therefore \angle CBP = 2\angle 1\))
\( \Rightarrow 2\angle 1 = {180^ \circ } – \angle B\)
\( \Rightarrow \angle 1 = {90^ \circ } – \frac{1}{2}\angle B…..\left({\text{i}} \right)\)

Again, \(\angle ACB\) and \(\angle QCB\) form a linear pair.
\(\therefore \angle ACB + \angle QCB = {180^ \circ }\)
\(\therefore \angle C + 2\angle 2 = {180^ \circ }\) (\(\because DC\) is the bisector of \(\angle QCB.\therefore \angle QCB = 2\angle 2\))
\( \Rightarrow 2\angle 2 = {180^ \circ } – \angle C\)
\( \Rightarrow \angle 2 = {90^ \circ } – \frac{1}{2}\angle C…..\left({{\text{ii}}} \right)\)

In \(\Delta BOC,\) we have
\(\angle 1 + \angle 2 + \angle BOC = {180^ \circ }\)
\( \Rightarrow {90^ \circ } – \frac{1}{2}\angle B + {90^ \circ } – \frac{1}{2}\angle C + \angle BOC = {180^ \circ }\) (Using \(\left({\text{i}} \right)\) and \(\left({\text{ii}} \right)\))
\( \Rightarrow {180^ \circ } – \frac{1}{2}\left({\angle B + \angle C} \right)\angle BOC = {180^ \circ }\)
\( \Rightarrow \angle BOC = \frac{1}{2}\left({\angle B + \angle C} \right)\)
\( \Rightarrow \angle BOC = \frac{1}{2}\left({{{180}^ \circ } – \angle A} \right)\,\left({\angle A + \angle B + \angle C = {{180}^ \circ } \Rightarrow \angle B + \angle C = {{180}^ \circ } – \angle A} \right)\)
\( \Rightarrow \angle BOC = {90^ \circ } – \frac{1}{2}\angle A\)

The sum of exterior angles of a triangle is \({360^ \circ }.\)

From the above figure, the exterior angles \(A\) and \(B\) make up a linear pair.

Therefore, \(\angle A + \angle B = {180^ \circ } \Rightarrow \angle A = {180^ \circ } – \angle B\)

The sum of all three exterior angles of the above triangle is given by

\(\angle A + \angle A + \angle A ={180^ \circ } – \angle B + {180^ \circ } – \angle B+ {180^ \circ } – \angle B\)

\( \Rightarrow \angle A + \angle A + \angle A + \angle B + \angle B + \angle B = {180^ \circ } + {180^ \circ } + {180^ \circ }\)

\( \Rightarrow \angle A + \angle B + \angle A + \angle B + \angle A + \angle B = {180^ \circ } + {180^ \circ } + {180^ \circ }\)

\( \Rightarrow 3\angle A + 3\angle B = {540^ \circ }……\left({\text{i}} \right)\)

Also, we know that the sum of interior angles of a triangle is \({180^ \circ }.\)

\( \Rightarrow \angle B + \angle B + \angle B = {180^ \circ }\)

\( \Rightarrow 3\angle B = {180^ \circ }……\left({{\text{ii}}} \right)\)

Substituting equation \(\left({{\text{ii}}}\right)\) in equation \(\left({{\text{i}}}\right),\) we get

\( \Rightarrow 3\angle A + {180^ \circ } = {540^ \circ }\)

\( \Rightarrow 3\angle A = {540^ \circ } – {180^ \circ }.\)

\( \Rightarrow 3\angle A = {360^ \circ }\)

\( \Rightarrow \angle A + \angle A + \angle A = {360^ \circ }\)

Therefore, the sum of exterior angles in a triangle is \({360^ \circ }.\)

Below are the properties of an exterior angle of a triangle.

1. The sum of the opposite internal angles of a triangle equals the triangle’s exterior angle.

2. In all circumstances, if an equivalent angle is taken at each vertex of the triangle, the exterior angles sum up to \({360^ \circ }.\) Indeed, this statement applies to any convex polygon, not only triangles.

3. According to the external angle theorem, the measure of an exterior angle is equal to the sum of the two distant interior angles of a triangle.

4. According to the external angle inequality theorem, the measure of any exterior angle of a triangle is greater than either of the opposite interior angles.

5. The adjacent interior angle and the exterior angle are supplementary. The total of a triangle’s outer angles is \({360^ \circ }.\)

Q.1. In the given triangle \(\Delta PQR,\) find the sum of \(a + b + c + x + y + z.\)

Ans: We know that in a triangle, the sum of exterior angles is equal to \({360^ \circ }.\)
From the given triangle, \(\Delta PQR,a,b,c\) form a set of exterior angles and \(x,y,z\) form another set of exterior angles.
So, \(a + b + c + x + y + z = {360^ \circ } + {360^ \circ } = {720^ \circ }\)

Q.2. In the below figure, sides \(QP\) and \(RQ\) of \(\Delta PQR\) are produced to point \(S\) and \(T\) respectively. If \(\angle SPR = {135^ \circ }\) and \(\angle PQT = {110^ \circ },\) find \(\angle PRQ.\)

Ans: Since \(QPS\) is a straight line.
\(\angle QPR + \angle SPR = {180^ \circ }\)
\( \Rightarrow \angle QPR + {135^ \circ } ={180^ \circ }\)
\( \Rightarrow \angle QPR = {180^ \circ } – {135^ \circ } = {45^ \circ }\)
Using Exterior angle property in \(\Delta PQR,\) we have
\(\angle PQT = \angle QPR + \angle PRQ\)
\( \Rightarrow {110^ \circ } ={45^ \circ } + PRQ\) \( \Rightarrow \angle PRQ = {110^ \circ } – {45^ \circ } = {65^ \circ }\)
Therefore, the measure of \(\angle PRQ = {65^ \circ }\)

Q.3. In the below figure, if \(QT \bot PR,\angle TQR = {40^ \circ }\) and \(\angle SPR = {30^ \circ },\)find \(x\) and \(y.\)
Ans: In \(\Delta TQR,\) we know two angles \(\angle TQR\) and \(\angle QTR.\) Therefore, we apply the angle sum property in \(\Delta TQR.\)

\( \Rightarrow \angle TQR + \angle QTR + \angle TRQ = {180^ \circ }\)
\( \Rightarrow {40^ \circ } + {90^ \circ } + \angle TRQ = {180^ \circ }\)
\( \Rightarrow \angle TRQ = {180^ \circ } – {130^ \circ } = {50^ \circ }\)
\( \Rightarrow x = {50^ \circ }\)

In \(\Delta PSR,\) using exterior angle property, we have
\(\angle PSQ = \angle PRS + \angle RPS\)
\( \Rightarrow y = x + {30^ \circ }\)
\( \Rightarrow y = {50^ \circ } + {30^ \circ }\)
\( \Rightarrow y = {80^ \circ }\)
Therefore, the values of \(x = {50^ \circ }\) and \(y = {80^ \circ }\)

Q.4. An exterior angle of a triangle is \({110^ \circ },\) and one of its interior opposite angles is \({30^ \circ }.\) Find the other two angles of the triangle.
Ans: Let \(ABC\) be a triangle whose side \(BC\) is produced to form an exterior angle \(ACD\) such that ext. \(\angle ACD = {110^ \circ }\)

Let \(\angle B = {30^ \circ }.\) By exterior angle theorem, we have
ext. \(\angle ACD = \angle B + \angle A\)
\( \Rightarrow {110^ \circ } = {30^ \circ } + \angle A\)
\( \Rightarrow \angle A ={110^ \circ } – {30^ \circ }\)
\( \Rightarrow \angle A = {80^ \circ }\)

In \(\Delta ABC,\) we have
\(\angle A + \angle B + \angle C = {180^ \circ }\)
\( \Rightarrow {80^ \circ } + {30^ \circ } + \angle C = {180^ \circ }\)
\( \Rightarrow \angle C = {180^ \circ } – {80^ \circ } – {30^ \circ }\)
\( \Rightarrow \angle C = {180^ \circ } – {110^ \circ }\)
\( \Rightarrow \angle C = {70^ \circ }\)
Therefore, the other two angles of the triangle are \({80^ \circ }\) and \({70^ \circ }.\)

Q.5. The sides \(BC,CA\) and \(AB\) of a \(\Delta ABC,\) are produced in order, forming exterior angles \(\angle ACD,\angle BAE\) and \(\angle CBF.\) Find the sum of \(\angle ACD + \angle BAE + \angle CBF.\)

Ans: In \(\Delta ABC,\) by using the exterior angle theorem, we have
\(\angle ACD = \angle 1 + \angle 2\)
\(\angle BAE = \angle 2 + \angle 3\)
\(\angle CBF = \angle 1 + \angle 3\)

Adding these three, we get
\(\angle ACD + \angle BAE + \angle CBF = \angle 1 + \angle 2 + \angle 2 + \angle 3\angle 1 + \angle 3\)
\( \Rightarrow \angle ACD + \angle BAE + \angle CBF = 2\left({\angle 1 + \angle 2 + \angle 3} \right)\)
\( \Rightarrow \angle ACD + \angle BAE + \angle CBF = 2 \times {180^ \circ } – {110^ \circ } = {360^ \circ }\)(Because, \(\angle 1 + \angle 2 + \angle 3 = {180^ \circ }\))

In this article, we have studied the meaning of linear pair of angles and interior opposite angles. Also, we have learned the definition of the exterior angle of a triangle with examples and properties of the exterior angle of a triangle. Also, we have determined the sum of exterior angles in a triangle and proved the theorems based on the exterior angles of a triangle and solved some example problems on the exterior angle theorem.

Q.1. How do you find the exterior angle of a triangle?
Ans: The sum of its two opposite non-adjacent interior angles is equal to the exterior angle of a triangle.
Formula to find the measure of an exterior angle of an equilateral triangle is \(\frac{{{{360}^ \circ }}}{n}\) Where \(n\) is the number of sides.

Q.2. What is the exterior angle of a triangle?
Ans: The angle formed by one side of a triangle and the extension of an adjacent side is called an exterior (or external) angle.

Q.3. What is the sum of the three exterior angles of a triangle?
Ans: The sum of three exterior angles in a triangle is \({360^ \circ }\)

Q.4. State exterior angle inequality theorem.
Ans: The sides \(AB\) and \(AC\) of a \(\Delta ABC\) are produced to \(P\) and \(Q\) respectively. If the bisectors of \(\angle PBC\) and \(\angle QCB\) intersect at \(O,\) then \(\angle BOC = {90^ \circ } – \frac{1}{2}\angle A\)

Q.5. State the exterior angle theorem.
Ans: If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.
Corollary: An exterior angle of a triangle is greater than either of the interior opposite angles.

Now you are provided with all the necessary information on the exterior angle of a triangle and its property and we hope this detailed article is helpful to you. If you have any queries regarding this article, please ping us through the comment section below and we will get back to you as soon as possible.

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