## ANGLE SUM PROPERTY

**INTRODUCTION: ** A plane figure formed by three line segments which are non-parallel to each other is called **triangle.**

If A,B,C are three non-collinear points in the plane of the paper, then the figure made up by the three line segments AB, BC and CA is called a triangle with vertices A,B,C. The triangle contains three vertices A,B and C and three non-parallel line segments AB,BC and CA. This triangle is denoted by Δ**ABC. ** AB, BC and CA are sides of **ΔABC. **Three angles are ˂BAC, <ABC and <ACB.

**ELEMENTS **or ** PARTS: **The three sides are **AB, BC, CA **and three angles **<A, <B, <C** of ΔABC are together called the six parts or elements of **ΔABC.**

**INTERIOR **and **EXTERIOR **of ** TRIANGLE: **We observe that all points in the plane ΔABC are divided into following three parts:

**INTERIOR :**The part made up by all such points**P**which are enclosed by**ΔABC**is called*the interior of*ΔABC.**EXTERIOR:**The part made up by all such point**Q**which are not enclosed by**ΔABC**is called**the exterior of ΔABC.****TRIANGULAR REGION:**The interior of**ΔABC**itself includes**P**and**R**is called**the triangular region ΔABC.**

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** TYPES OF TRIANGLES**

- Naming of triangles by considering the lengths of their sides:-
**Scalene Triangle:**

A triangle whose no two sides are equal is called a **scalene triangle. ΔABC **is a scalene triangle.

**Isosceles Triangle:**

A triangle whose two sides are equal is called an **isosceles triangle. **Δ**ABC **is an isosceles triangle so **AB = AC.**

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**Equilateral Triangle:**

A triangle whose all sides are equal to one another, is called an **equilateral triangle. ΔABC **is an equilateral triangle where **AB = AC = BC.**

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- Naming triangles by considering the measures of their angles:-
**Acute Triangle :**

A triangle whose all the angles are acute is called an **acute angled triangle **or an **acute triangle.**

**ΔABC **is an acute-angled triangle where <A, <B and < C are acute angles. Equilateral triangle is an acute-angled triangle because the measure of its each angle is 60°.

**Right Triangle:**

A triangle whose one angle is a right angle, is called **right-angled triangle** or a ** right triangle.**

The side opposite to the right triangle is known as **hypotenuse ** and other two sides are called the **legs of the triangles.**

**ΔABC ** is a right-angled triangle, where **<B = 90° **remaining two angles are acute.

**Obtuse triangle:**

A triangle whose one angle is obtuse, is called an **obtuse-angled triangle **or **an obtuse triangle, **where <E is an obtuse angle.

** ANGLE SUM PROPERTY OF A TRIANGLE**

** PROPERTY: ** The sum of the angles of a triangle is two right angles or 180°

**PROOF: **Let ΔABC be any triangle. Through A, a line XY parallel to the side BC . Since, XY||BC and the and the transversal AB cuts XY and BC at A and B respectively.

**<1 = <4 [ as alternate interior angles are equal ]**

Similarly , XY ||BC and the transversal AC cuts XY and BC at A and C respectively.

**<2 = <5 [ as alternate interior angles are equal]**

** **Also ** <3 = <3**

Adding the angles on the respective sides, we get

**<1 + <2 + <3 = <4 + <5 + <3**

But, **<4 + <5 + <3 = 180° = 2 right angles.**

Hence, the sum of the angles of a triangle is two right angles or 180°.

From the above property, we obtain the following useful results:-

**A triangle cannot have more than one right angle.**

*A triangle cannot have more than one obtuse angle i.e if one angle of the triangle is obtuse, then other two are acute.*

**In a right triangle, the other two angles are always acute and their sum is 90°.**

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** EXTERIOR ANGLE PROPERTY OF A TRIANGLE**

**PROPERTY: ** When a side of a triangle is produced then the exterior angle so formed is equal to the sum of its interior opposite angle.

To prove, in **ΔABC, ** BC has been produced to point D, forming an exterior angle <ACD.

**To prove : <ACD = <CAB + < ABC**

*CONSTRUCTION: ** *from C, CE is drawn which parallel to BA.

** PROOF:** Since CE||BA and AC is transversal.

**So we have <ACE = < CAB [ alternate angle]**

Again, **CE ||BA ** and BCD is a transverse

**So, <ECD = <ABC **[**corresponding angle]**

**Adding the corresponding sides**

**<ACE + <ECD = <CAB + <ABC**

**<ACD = <CAB + < ABC**

Therefore, it is proved that exterior angle of a triangle is always equal to the sum of its interior opposite angle.

Let’s discuss some problem on the angle sum property of triangle:

**Problem 1: Two angles of a triangle measures 63° and 47° respectively. Find the measure of the third angle of the triangle.**

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**Solution: ** Let the measure of the third angle be x°; since the sum of all the three angles of a triangle is 180°, we have

63 + 47 + x = 180

Or, 110 + x = 180

Or, x = 180 – 110 = 70

*Therefore, the measure of the third angle is 70°*

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**Problem 2: The angles of a triangle are in the ratio 2: 3: 4, find the measure of each angle of the triangle.**

**Solution: ** Let ** **the measures of the given angles be **2x° , 3x° , 4x°,**

Since the sum of the angles of a triangle is **180°**, so we have : **2x + 3x + 4x = 180**

Or, 9x = 180

Or, x = 180/9 = 20

Angles are : 2 X 20 = 40

3 X 20 = 60

4 X 20 = 80

*Therefore, the measures of three angles are 40° , 60° , 80°.*

**Problem 3: Calculate the value of x in the downward figure.**

**Solution: ** In ΔABC, we have:

<BAC + <ABC + <BCA = 180°

Or, 40°+ <ABC + 90 = 180°

Or, <ABC + 130° = 180°

Or, <ABC = 180 – 130 = 50°

IN ΔBDE , we have:

<EBD + <BDE + < DEB = 180°

Or, <ABC + <BDE + < DEB = 180°

Or, 50° + x° + 100° = 180°

Or, x° + 150°= 180°

Or, x° = 180 – 150 = 30°

*Therefore, the value of x = <DEB = 30°.*

**Problem 4 : Calculate the value of x in the downward figure.**

**Solution: **AD is joined to produce to E.

**Since the exterior angle of a triangle is equal to the sum of interior opposite angles, so**

In ΔACD, <CDE = <DCA + <CAD…………………………(i)

In ΔABD, < EDB = <DAB + <ABD………………………….(ii)

Adding (i) and (ii) , we have;

<CDE + <EDB = <DCA + <CAD + <DAB + <ABD

Or, x° = 30° + (<CAD + <DAB) + 45°

Or, x° = 30° + 55° + 45° = 130°

Or, x° = 130°

*Therefore, the value of x = 130°.*

**Problem 5: In a right triangle, one of the acute angles is 58°. Find the other acute angle.**

**Solution: **Let the measure of the other acute angle be x°, then the angles of the triangle are 90°, 58° and x° .

**Since sum of all three angles of triangle is 180****◦ according to the angle sum property,**

So, 90° + 58° + x° = 180°

Or, 148° + x° = 180°

Or, x = 180 – 148

Or, x = 32°

*Therefore, the measure of the other acute angle is 32°.*

**Problem 6: In the adjoining figure ΔABC is right-angled at <C, and CD ḻ AB** , **also <A = 65°. Find (i) ACD ; (ii) <BCD; (iii) <CBD.**

**Solution: **ΔABC is right-angled triangle,

<C = 90°

<A = 65°

So, <B = 180° – (90° + 65°)

Or, <B = 180 – 155 = 25°

Or, <CBD = 25°

As, CD ḻ AB, Therefore, <ADC = <CDB = 90°

In, ΔACD, we have,

<ACD + <CAD + <ADC = 180°

Or, <ACD + 65°+ 90° = 180°

Or, <ACD = 180 – (90 +65) = 25°

In ΔBCD, we have,

<BCD + <CBD + <BDC = 180°

Or, <BCD + 25° + 90° = 180°

Or, <BCD = 180 – ( 90 + 25) = 65°

*Therefore, <ACD = 25° , <BCD = 65° , <CBD = 25°.*

**Problem 7: In ΔABC, D,E are points on sides AB and AC in such a way that DE||BC. If <B = 30° and <A = 40°, find x°,y°, z°.**

**Solution: ** In ΔABC, we have,

<A = 40° and <B = 30°

So, <A + <B + <C = 180°

Or, 40°+ 30° + <C = 180°

Or, <C = 180° – (40° +30°)

Or, <C = 110°

DE||BC and transversal AB cuts them at E and c respectively.

<ACB = <AED ……………………. (corresponding angles)

So, y° = z° = 110°

Again, DE||BC and transversal AB cuts them at D and B respectively.

<B = <ADE ………………………….(corresponding angles)

x° = 30°

*Therefore, x° = 30° ; y° = 110° ; z° =110°.*

**Problem 8: This figure has been obtained by using two triangles. Find out <A+<B+<C+<D+<E+<F**

**Solution: **We know that *the sum of the angles of a triangle is 180 °*

In ΔACE, we have <A +<C+<E = 180°……………..(i)

In ΔBDF, we have <B+<D+<F = 180°………………(ii)

Adding the corresponding sides of (i) and (ii)

We get : <A+<C+<E+<B+<D+<F= 180° + 180°,

So, <A+<B+<C+<D+<E+<F = 360°

*Therefore, sum of the six angles are 360°.*

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**Problem 9 : Calculate the value of of x.**

**Solution: **DBC is a straight line, we have:

<ABD + <ABC = 180°

<ABC = 180 – (<ABD) = 180 – 138 = 42°.

BC = AC = < BAC = <ABC = 42°.

In ΔABC, we have;

<BAC +<ABC +<BCA = 180°

Or, 42° + 42° + x° = 180°

Or, 84° + x° = 180°

Or, x° = 180 – 84 = 96°

*Therefore, value of x = 96°*

**Problem 10: Calculate the value of x.**

**Solution: **AB = AC ; so, <ACB = <ABC ………(angles opposite to equal sides of a triangle are equal)

Now, in ΔABC, we have;

<BAC + <ABC +<ACB = 180°

Or, 90°+ 2<ABC = 180°

Or, 2<ABC = 180 – 90

So, <ABC = <BCA = 45°

AC = CD

<CAD = <ADC = x°

Since, *exterior angle of a triangle is equal to the sum of interior opposite angles,*

So we have <ACB = <CAD +<ADC = 45°……..( <ACB = x° + x° )

2x° = 45°

Or, x° = (22^{1}/_{2})°

*Therefore, the value of x = (22 ^{1}/_{2})°. *

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