## 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.**

** **

** 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.**

** **

**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.**

** **

- 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°.**

** **

** 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.**

** **

**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°*

* *

* *

**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°.*

* *

**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})°. *

* *

* *

* *

* *

* *

* *

it worth a lot more than they sell for.

But wanna input on few general things, The website design and style is perfect, the subject material is very fantastic : D. Ursula Nathaniel Berte

As soon as I found this website I went on reddit to share some of the love with them. Misti Leonardo Kryska

If you desire to take much from this piece of writing then you have to apply these techniques to your won weblog. Elbertina Lombard Fairlie

This is a great addition to the photos you posted before. Deina Eldridge Ellersick

This is my first time pay a visit at here and i am actually pleassant to read everthing at single place. Alvira Pattie Mayne

I am actually thankful to the owner of this web site who has shared this fantastic post at at this time. Hyacinth Garvy Plunkett

Hi there friends, pleasant piece of writing and nice urging commented here, I am truly enjoying by these. Annadiana Byron Mona

I see something truly interesting about your site so I bookmarked . Lana Lorry Jacobs

Hello there. I found your web site via Google while looking for a comparable matter, your web site got here up. It looks good. I have bookmarked it in my google bookmarks to visit then. Lou Hurley Valdas

Excellent article. I certainly appreciate this website. Keep writing! Austin Lewes Griff

I wish to voice my admiration for your generosity for folks who actually need help with the matter. Your personal commitment to passing the message all-around was extraordinarily invaluable and has regularly helped somebody just like me to attain their pursuits. This valuable guide means a great deal a person like me and extremely more to my peers. Thank you; from all of us. Sibylle Reggie Devine

Hi, I think your site might be having browser compatibility issues. When I look at your website in Firefox, it looks fine but when opening in Internet Explorer, it has some overlapping. I just wanted to give you a quick heads up! Other then that, amazing blog! Alexis Cass Janot

Wonderful photos, Em! Thank you for sharing them with me. Love,Betty Bryana Payton Hellman

Hi, I think your blog may be having web browser compatibility issues. When I look at your web site in Safari, it looks fine however, if opening in Internet Explorer, it has some overlapping issues. I simply wanted to provide you with a quick heads up! Aside from that, great site! Aurelie Web Godart

Awesome write-up. I am a normal visitor of your web site and appreciate you taking the time to maintain the nice site. I will be a frequent visitor for a really long time. Jinny Nikolaos Suzie