Relations among Sides and Angles of a Triangle: It is a fact that, as the name suggests, a triangle has three angles and three sides. Mathematically, a triangle is defined as any closed figure that is formed by three intersecting lines. Although triangle properties such as sides, angles, bisectors, congruence, and similarity are more commonly used, how do you think the sides and angles in a triangle are related to each other? Do you think any three-line segments can form a triangle? In a triangle, the measure of sides can often be used to calculate the angles and vice versa.

In this article, let us learn how the sides and angles of triangles are related and learn theorems that deal with this relationship.

Relations Between Sides of a Triangle

The triangle inequality theorem defines the relationship between the sides of a triangle. The theorem states that the sum of lengths of two sides of a triangle is greater than the length of the third side.

In the shown figure, the following inequalities hold.

\(a+b>c\)
\(b+c>a\)
\(c+a>b\)

This theorem also serves as a condition to check if triangles can be formed using the side lengths provided. In other words, if and only if all these three inequalities are true, the given side lengths form the triangle.

Types of Angles in a Triangle

In any triangle, there are three sides, and there are two types of angles, viz. interior angles, and exterior angles, and they are identified as shown in the figure below.

Observe that the exterior angles are supplementary to the interior angles in the triangle.

For example, 

1. If \(\angle {\rm{a = 5}}{{\rm{0}}^{\rm{o}}},\) then its corresponding exterior angle is \({\rm{ = 13}}{{\rm{0}}^{\rm{o}}}\)
2. If \(\angle b{\rm{ = 10}}{{\rm{0}}^{\rm{o}}},\) then its corresponding exterior angle is \({\rm{ = 8}}{{\rm{0}}^{\rm{o}}}\)
3. If \(\angle a{\rm{ = 3}}{{\rm{0}}^{\rm{o}}},\) then its corresponding exterior angle is \({\rm{=\,15}}{{\rm{0}}^{\rm{o}}}\)

Relations Between Angles of a Triangle

The sum of all the angles in a triangle is \({\rm{18}}{{\rm{0}}^{\rm{o}}}.\) Consider the triangle shown below.

Here, \(A + B + C = {\rm{18}}{{\rm{0}}^{\rm{o}}}.\)

Classification of Triangles

There are various tools to discover the sides and angles in triangles. They are broadly classified into two types:
1. Geometric relationship between sides and angles
2. Trigonometric relationship between sides and angles

Geometric Relationship Between Sides and Interior Angles

The angle-side relationship theorem defines the geometric relation between sides and interior angles. 

Angle-side relationship theorem states that in any triangle:
1. The smallest angle is opposite to the smallest side
2. The largest angle is opposite to the largest side
3. The second-largest angle is opposite to the second-largest side

Right Triangles: Pythagorean Theorem

Pythagoras’ theorem states that, in a right triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides. 

\({\rm{hypotenus}}{{\rm{e}}^{\rm{2}}}{\rm{ = perpendicula}}{{\rm{r}}^{\rm{2}}}{\rm{ + bas}}{{\rm{e}}^{\rm{2}}}\)

In the given triangle, 

\(a?\) perpendicular
\(b?\) base
\(c?\) hypotenuse

\(\therefore {c^2} = {a^2} + {b^2}\)

Trigonometric Relationship of Right Triangles

For \(\theta ,\) an acute angle in a right triangle, the trigonometric functions are defined as follows.

\( \sin \theta = \frac{{{\rm{ opposite }}}}{{{\rm{ hypotenuse }}}}\)

\(\cos \,\theta = \frac{{{\rm{adjacent}}}}{{{\rm{hypotenuse}}}}\)

\(\tan \,\theta = \frac{{{\rm{opposite}}}}{{{\rm{adjacent}}}}\)

\( \cot \theta = \frac{{{\rm{ adjacent }}}}{{{\rm{ opposite }}}}\)

\( \sec \theta = \frac{{{\rm{ hypotenuse }}}}{{{\rm{ adjacent }}}}\)

\({\rm{cosec}}\,\theta = \frac{{{\rm{hypotenuse}}}}{{{\rm{opposite}}}}\)

Trigonometric Relationship of Oblique Triangles

An oblique triangle is any triangle that is not a right triangle.

Two formulas help us solve such oblique triangles.

1. Law of Sines

Also called the sine rule, this law states that the ratio of the length of a side and the angle opposite to that side is a constant for all the sides and angles in a triangle.

It is given by,

\(\frac{a}{{ \sin A}} = \frac{b}{{ \sin B}} = \frac{c}{{\sin C}}\)

Proof: From the figure, h is the altitude. For each triangle, we know that, 

\(\sin A = \frac{h}{b}\)

\(\sin B = \frac{h}{a}\)

Rearranging, we get, \(h = a \sin B.\)

\( \Rightarrow \sin A = \frac{{a\sin B}}{b}\)

\(\frac{a}{{\sin A}} = \frac{b}{{ \sin B}}\)

Similarly, we can also write,

\(\frac{b}{{ \sin B}} = \frac{c}{{ \sin C}}\)

Equating, we get,

\(\frac{a}{{ \sin A}} = \frac{b}{{ \sin B}} = \frac{c}{{ \sin C}}\)

Note that in order to use the law of sines, you have to know either two angles and a side length or two side lengths and an angle that is opposite to one of them.

2. Law of Cosines 

Law of cosines is used when lengths of two sides and an included angle or the lengths of three sides are known. It states that,

\({a^2} = {b^2} + {c^2} – 2bc \cos A\)

\({b^2} = {c^2} + {a^2} – 2ca \cos B\)

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

Proof: From the figures, we can write,

\({h^2} = {a^2} – {x^2}\)

\({b^2} = {h^2} + {(c + x)^2}\)

Substituting the value of \(h,\) we get,

\({b^2} = {a^2} – {x^2} + {c^2} + 2cx + {x^2}\)

\({b^2} = {a^2} + 2cx + {c^2}\)

From the figure, \(x = a \cos \left( {{{180}^{\rm{o}}} – B} \right) =\, – a\cos \cos B\)

\(\therefore {b^2} = {a^2} + {c^2} – 2ca \cos B\)

Using similar arguments, we can derive the other two formulas too.

Observe that this is similar to the Pythagorean Theorem, except that, in a right triangle, \(\angle C = {90^{\rm{o}}},\) and \(\cos \, {90^{\rm{o}}} = 0.\) Hence, there will be no third term.

The Pythagorean theorem can be considered a special case of the Law of Cosines.

3. Law of Tangents

The law of tangents establishes the relationship between two sides of a triangle and the tangents of sum and difference of the opposite angles. It is defined as,

\(\tan \frac{{B – C}}{2} = \frac{{b – c}}{{b + c}}\cot \frac{A}{2}\)

\(\tan \frac{{C – A}}{2} = \frac{{c – a}}{{c + a}}\cot \frac{B}{2}\)

\(\tan \frac{{A – B}}{2} = \frac{{a – b}}{{a + b}}\cot \frac{C}{2}\)

Proof: Using the law of sines,

\(\frac{a}{{ \sin A}} = \frac{b}{{ \sin B}} = \frac{c}{{ \sin C}} = k(say)\)

\(\therefore \frac{{b – c}}{{b + c}} = \frac{{k( \sin B – \sin C)}}{{k( \sin B + \sin C)}}\)

\( = \frac{{2 \cos \frac{{B + C}}{2} \sin \frac{{B – C}}{2}}}{{2 \sin \frac{{B + C}}{2} \cos \frac{{B – C}}{2}}}\)

\(\cot \frac{{B + C}}{2} \tan \frac{{B – C}}{2}\)

\( = \cot \left( {\frac{\pi }{2} – \frac{A}{2}} \right) \tan \frac{{B – C}}{2}\)

\( = \frac{{ \tan \frac{{B – C}}{2}}}{{ \cot \frac{A}{2}}}\)

\(\therefore \tan \frac{{B – C}}{2} = \frac{{b – c}}{{b + c}} \cot \frac{A}{2}\)

Using similar arguments, we can derive the other two formulas too.

4. Projection Rule

The triangles have \(6\) parameters. The projection formula expresses the sum of two sides in terms of the third side.

Projection law states that in any triangle,

\(b\cos C + c\cos B = a\)

\(a\cos C + c \cos A = b\)

\(a\cos B + b\cos A = c\)

Proof: Using the law of sines,

\(\frac{a}{{ \sin A}} = \frac{b}{{ \sin B}} = \frac{c}{{ \sin C}} = k\,(say)\)

Rearranging, we get,

\(a = k\sin A\)

\(b = k\sin B\)

\(c = k\sin C\)

To prove \(b\cos C + c\cos B = a,\)

Substituting the values of \(b\) and \(c,\) we get,

L.H.S. \( = k\sin B\cos C + k\sin C \cos B\)

\( = k( \sin B \cos C + \sin C \cos B)\)

\( = k \sin (B + C)\)

In any triangle, \(A + B + C = \pi ,\)

\( = k \sin (\pi – A)\)

\( = k \sin A\)

\(=a?\) R.H.S.

Using similar arguments, we can derive the other two formulas too.

5. m-n Theorem

For a point \(D\) on \(BC\) that divides it in the ratio \(m:n,\) the theorem states that,

\((m + n) \cot \theta = m \cot \alpha – n \cot \beta \)

\((m + n) \cot \theta = m \cot B – n \cot C\)

Proof:

Given: \(\frac{{BD}}{{DC}} = \frac{m}{n}\) and \(\angle ADC = \theta \)

\(\angle ADB = {180^{\rm{o}}} – \theta \)

\(\angle BAD = \alpha \)

\(\angle DAC = \beta \)

So, \(\angle ABD = \theta – \alpha = B,\) and \(C = {180^{\rm{o}}} – (\theta + \beta )\)

In \(\Delta ABD,\frac{{BD}}{{ \sin \alpha }} = \frac{{AD}}{{ \sin (\theta – \alpha )}}\)

In \(\Delta ADC,\frac{{DC}}{{ \sin \beta }} = \frac{{AD}}{{ \sin (\theta + \beta )}}\)

\(\frac{{BD}}{{DC}}\frac{{ \sin \beta }}{{ \sin \alpha }} = \frac{{ \sin (\theta + \beta )}}{{ \sin (\theta – \alpha )}}\)

\( \Rightarrow \frac{{m \sin \beta }}{{n \sin \alpha }} = \frac{{ \sin (\theta + \beta )}}{{ \sin (\theta – \alpha )}}\)

\(\frac{{m \sin \beta }}{{n \sin \alpha }} = \frac{{ \sin \theta \cos \beta + \cos \theta \sin \beta }}{{ \sin \theta \cos \alpha – \cos \theta \sin \alpha }}\)

\(m \sin \beta ( \sin \theta \cos \alpha – \cos \theta \sin \alpha ) = n \sin \alpha ( \sin \theta \cos \beta + \cos \theta \sin \beta )\)

\(m \cot \alpha – m \cot \theta = n \cot \beta + n \cot \theta \)

\((m + n) \cot \theta = m \cot \alpha – n \cot \beta \)

Similarly, we can get the other argument too.

Solved Problems – Relations Among Sides and Angles of a Triangle

Q.1. Devin is making a garden in his yard. He wants to check if the angle at \(C\) is a right angle. The dimensions are as marked in the diagram.

Help Devin identify the type of angle at \(C.\)
Ans: The side lengths of a right triangle satisfy the Pythagorean theorem.
\({\rm{hypotenus}}{{\rm{e}}^{\rm{2}}}{\rm{ = perpendicula}}{{\rm{r}}^{\rm{2}}}{\rm{ + bas}}{{\rm{e}}^{\rm{2}}}\)
In Devin’s yard,
perpendicular ? 52 m
base ? 48 m
hypotenuse ? 20 m
\(\therefore {52^2} = {20^2} + {48^2}\)
\(2704=400+2304\)
\(2704=2704\)
Hence, \(C\) is a right angle.

Q.2. Compare the lengths of the sides of the triangle given below.

Ans: Sum of interior angles in a triangle \( = {180^{\rm{o}}}\)
\( \Rightarrow \angle E + \angle F + \angle G = {180^{\rm{o}}}\)
\(\angle E + {30^{\rm{o}}} + {65^{\rm{o}}} = {180^{\rm{o}}}\)
\(\angle E = {180^{\rm{o}}} – {95^{\rm{o}}}\)
\(\therefore \angle E = {85^{\rm{o}}}\)
Descending order of angles, \(\angle E > \angle G > \angle F\)
Using the side-angle relationship theorem,
Note: consider the side that is opposite to the angle in consideration.
∴ The comparison of sides can be written as \(\overline {FG} > \overline {EF} > \overline {GE} \)

Q.3. In a \(\Delta ABC,\angle A = {60^{\rm{o}}},\angle B = {80^{\rm{o}}},\) and \(BC = 7\;{\rm{cm}}\) Find \(AC.\)
Ans: Since we are given two angles and a side, we use the sine law.
asin \(\frac{a}{{\sin A}} = \frac{b}{{ \sin B}}\)
Here,
\(A = {60^{\rm{o}}}\)
\(B = {80^{\rm{o}}}\)
\(a = 7\;{\rm{cm}}\)
\(AC = b\)
\(\therefore \frac{7}{{ \sin {{60}^{\rm{o}}}}} = \frac{b}{{\sin {{80}^{\rm{o}}}}}\) (Note: Each fraction has a side and its opposite angle)
\( \Rightarrow b = \frac{7}{{\sin {{60}^{\rm{o}}}}} \times \sin {80^{\rm{o}}}\)
\(AC = b = 7.96\;{\rm{cm}}\)

Q.4. The sides of a triangle are \(5\;{\rm{cm}},7\;{\rm{cm}},\) and \(8\;{\rm{cm}}.\) Find the measure of the middle-sized angle.
Ans: The middle-sized angle in a triangle lies opposite the middle-sized side.
Therefore, the angle lies opposite to 7-cm side.
Cosine law is stated as, \({c^2} = {a^2} + {b^2} – 2ab \cos C\)
\(\therefore \cos A = \frac{{{5^2} + {8^2} – {7^2}}}{{2 \times 5 \times 8}}\)
\( \cos A = \frac{{25 + 64 – 49}}{{80}}\)
\( \cos A = \frac{{40}}{{80}} = 0.5\)
\( \Rightarrow A = {\cos ^{ – 1}}\left( {0.5} \right)\)
\(\therefore A = {60^{\rm{o}}}\)

Q.5. Prove using projection rule: \(a\left( {{b^2} + {c^2}} \right) \cos A + b\left( {{c^2} + {a^2}} \right) \cos B + c\left( {{a^2} + {b^2}} \right) \cos C = 3abc\)
Ans: L.H.S. \( \to a\left( {{b^2} + {c^2}} \right) \cos A + b\left( {{c^2} + {a^2}} \right) \cos B + c\left( {{a^2} + {b^2}} \right) \cos C\)
\( = a{b^2} \cos A + a{c^2} \cos A + b{c^2} \cos B + {a^2}b \cos B + {a^2}c \cos C + {b^2}c \cos C\)
\( = ab(b \cos A + a \cos B) + bc(c \cos B + b \cos C) + ac(c \cos A + a \cos C)\)
Using the projection formula, we have,
\( = ab(c) + bc(a) + ac(b)\)
\( = 3abc\)
Hence, proved.

Summary

This article helps us understand the various ways in which angles and sides of a triangle are related. It lists the side relations via the triangle inequality theorem, the angle relations via the sum of angles in a triangle. Then, it explains the geometric and trigonometric relations of sides and angles in a triangle. We also learn to prove the law of sines, law of cosines, law of tangents, projection rule, and \(m-n\) theorem.  The solved examples help us understand how to employ the thus learnt geometric and trigonometric relations among sides and angles of a triangle.

Frequently Asked Questions (FAQs)

Q.1. What are the relationships between side lengths and angle measures of triangles?
Ans: Angle-side relationship theorem states that in any triangle:
1. The smallest angle is opposite to the smallest side
The largest angle is opposite to the largest side
The second-largest angle is opposite to the second-largest side

Q.2. What is the relationship between the 3 sides of any triangle?
Ans: The sum of lengths of two sides in a triangle is greater than the length of the third side. This is called the triangle inequality theorem.

Q.3. Who established the relationship between sides and angles in a right-angled triangle?
Ans: Pythagorean theorem is named after the Greek philosopher and mathematician Pythagoras. Although the theorem may have been known \(1000\) years earlier, he was the first to prove it.

Q.4. What is the greatest number of right angles a triangle can contain?
Ans: The sum of angles in a triangle is \({180^{\rm{o}}}.\) If the measure of one angle is \({90^{\rm{o}}},\) then the sum of the other two angles will be \({90^{\rm{o}}}.\) This means that the measures of the other two are complementary angles, and they will be less than \({90^{\rm{o}}}\) each. Hence, a triangle can have a maximum of one right angle only.

Q.5. How do you find the angle of a right-angled triangle given two sides?
Ans: The angle measure can be calculated using the trigonometric relations between sides and angles of a right-angled triangle. For example, we can calculate the ratio of the opposite to hypotenuse as \(\sin \,\theta .\) Then, we find the of the value to get the measure of the angle.

We hope this detailed article on relations among sides and angles of a triangle helped you in your studies. If you have any doubts, queries or suggestions regarding this article, feel to ask us in the comment section and we will be more than happy to assist you. Happy learning!