**A plane figure bounded by three non concurrent line segment in plane is called a TRIANGLE.**

Clearly, when we say a plane figure, we actually mean the linear figure, not the two dimensional figure. Let A, B, C are non linear points such that they are not on the same line; we say A, B, C are ** noncollinear.** Join AB, BC , CA. We get a linear figure which consists of three line segments which meet only at their end points. Such a linear figure is called a triangle. We say ** A, B **and **C **are the **vertices** of the triangle ABC. The segments AB, BC and CA are called the **sides **of the triangle ABC; and the angles ∠ABC, ∠BCA, ∠CAB are the three angles of a triangle ABC.

A triangle consists of three elements:

Vertices | Sides | Angles |

A | AB | ∠BAC or ∠A |

B | BC | ∠ABC or ∠B |

C | CA | ∠BCA or ∠C |

Triangles are classified based on the measure of sides and angles of the triangle.

**Classification based on sides of the triangle:**

There are three types:

(i) Equilateral triangle

(ii) Isosceles triangle

(iii) Scalene triangle

Now let us study in brief:

**Classification based on angles of the triangle:**

There 3 types of triangle:

(i) Acute angled triangle

(ii) Right angled triangle

(iii) Obtuse angled triangle

**Exercise 3.2.1**

**1. Match the following:**

(i)

(a) Equilateral triangle

(ii)

(b) Acute angled triangle

(iii)

(c) Right angled triangle

(iv)

(d) Obtuse angled triangle

**Solution:**

————— Right angled triangle

—————– Obtuse angled triangle

——————- equilateral triangle

——————— Acute angled triangle

2. **Based the sides, classify the following triangles:**

(i)

**Solution:**

Scalene triangle [all the three sides are different from one another]

(ii)

**Solution:**

Scalene triangle [ all the three sides are different from one another]

(iii)

**Solution:**

Scalene triangle [ all the three sides are different from one another]

(iv)

**Solution:**

Isosceles triangle [ the two sides of triangle are equal to one another]

(v)

**Solution:**

Scalene triangle [ all the three sides are different from one another]

(vi)

**Solution:**

Scalene triangle [ all the three sides are different from one another]

(vii)

**Solution:**

Scalene triangle [ all the three sides are different from one another]

(viii)

**Solution:**

Equilateral triangle [ all the three sides are equal to one another]

(ix)

**Solution:**

Isosceles triangle [ the two sides of triangle are equal to one another]

(x)

**Solution:**

Isosceles triangle [ the two sides are triangle are equal to one another]

**3.2.2 Sum of interior angles**

**Theorem 1: **** In any triangle, the sum of the three interior angles is 180°**

This theorem is also called interior angle theorem.

**Given:** ABC is a triangle.

**To prove:** ∠ABC + ∠CAB + ∠ACB = 180**°**

**Construction:** through point A draw the line EF||BC

**Proof: **

Below we give several statements and the reason for the truth of each statement. Finally we arrive at the desired conclusion.

Statement:

∠ABC = ∠EAB [alternate angles by the transversal AB with the parallel lines BC and EF]

∠BCA= ∠FAC [alternate angles by the transversal AB with the parallel lines BC and EF]

∠EAB + ∠BAC + ∠FAC = 180 **° **[Sum of all linear angles at A]

By substituting ∠EAB = ∠ABC and ∠FAC = ∠BCA, we finally get,

∠ABC + ∠CAB + ∠ACB = 180°

This completes the proof.

**Example 1: **** In a triangle ABC, it is given that **∠B = 105**° and **∠C = 50**°. ****Find **∠A.

**Solution:**

We have, in triangle ABC, [by theorem 1]

∠A + ∠B + ∠C = 180**°**

We know,

∠B = 105° and ∠C = 50°

Therefore, ∠A = 180° – ∠B – ∠C

∠A = 180° – 105° – 50°

∠A = 25°

Thus, ∠A Measures 25°

**Example 2: ****In the figure, find all the angles:**

**Solution: **

**In triangle ABC, if we make use of theorem 1 [interior angle theorem], we get, **

∠A + ∠B + ∠C = 180° ——————– (1)

In the figure it is given that, ∠A = 5x , ∠B = 3x, ∠C = 2x

Let us substitute the value of ∠A, ∠B and ∠C in equation (1)

Then, we get,

5x + 3x + 2x = 180°

10x = 180°

Therefore, x = 18°

Thus, ∠A = 5x = 5 x 18° = 90°

∠B = 3x = 3 x 18° = 54°

∠C = 2x = 2 x 18° = 36°

** **

**Example 3: **If the bisectors of the angles ∠ABC and ∠ACB of a triangle meet at a point O, then prove that ∠BOC = 90**° + ½** ∠BAC

**Solution:**

**Given: **A triangle ABC and the bisectors of ∠ABC and ∠ACB meeting at point O.

**To prove: **∠BOC = 90**° + ½** ∠BAC

**Proof: **In triangle BOC, we have,

∠1 + ∠2 + ∠BOC = 180° ———-(1)

In triangle ABC, we have, ∠ABC + ∠CAB + ∠ACB = 180°

Since, BO and CO are bisectors of ∠ABC and ∠ACB respectively, we have,

∠B = 2∠1 and ∠C = 2∠2

Therefore, we get, ∠A + 2∠1 + 2∠2 = 180°

Let us divide it by 2, we get,

∠1 + ∠2 = 90° – ∠A /2 ——————-(2)

From (1) and (2), we get,

90° – ∠A /2 + ∠BOC = 180°

Hence, ∠BOC = 90° + ½ ∠BAC

__Exercise 3.2.2__

**In a triangle ABC, if**∠A = 55**° and**∠B = 40**°, find**∠C

Solution:

We know, ∠A + ∠B + ∠C = 180° ———– (1) [from theorem 1 – interior angle theorem]

It is given that,

∠A = 55° and ∠B = 40°

Substitute the value of ∠A and ∠B in equation (1)

Then,

55° + 40° + ∠C = 180°

∠C = 180° – 55° – 40°

∠C = 85°

** **

**In a right angled triangle, if one of the other two angles is 35****°****, find the remaining angle:**

Solution:

Data: triangle ABC is a right angled triangle and ∠C =35**°**

Now, we have find ∠C

We know, ∠A + ∠B + ∠C = 180**° —————–(1)**

Let us substitute the value of ∠A and ∠B in equation (1)

** **∠C = 180**°** – 90**° – **35**°**

∠C = 55**°**

Thus, ∠C measures 55°

3. If the vertex angle of an isosceles triangle is 50°, find the other angles.

Solution:

Data: Triangle ABC is an isosceles triangle, AB = AC and ∠A = 50**°**

We know, AB = AC, therefore ∠B = ∠C

We know, ∠A + ∠B + ∠C = 180**° ———– (1) ****[from theorem 1 – interior angle theorem]**

∠A = 50°

Substitute the value of ∠A in equation (1)

Then,

50**° ****+** ∠B** + **∠C = 180**°**

We know, ∠B = ∠C

∠B + ∠B = 180**° – **50**° **

2∠B = 130**°**

∠B = 130°/2 = 65°

∠B = ∠C = 65**°**

**4. The angles of a triangle are in the ratio 1:2:3. Determine the three angles.**

**Solution:**

Let ∠A = 1x, ∠B = 2x, ∠C = 3x

Then,

∠A + ∠B + ∠C = 180° ———– (1) [from theorem 1 – interior angle theorem]

Substitute the value of ∠A, ∠B and ∠C in equation (1)

Then,

x + 2x + 3x = 180°

6x = 180°

x = 180°/6

x = 30°

Therefore,

∠A = 30°,

∠B = 2x = 2 x 30 = 60°,

∠C = 3x = 3 x 30 = 90°

5. In adjacent triangle ABC, find the value of x and calculate the measure of all the angles of the triangle.

Let ∠A = x + 15, ∠B = x – 15, ∠C = x+30

Then,

∠A + ∠B + ∠C = 180° ———– (1) [from theorem 1 – interior angle theorem]

Substitute the value of ∠A, ∠B and ∠C in equation (1)

Then,

(x + 15) + (x -15) + (x + 30) = 180°

3x + 30 = 180°

3x = 150

x = 150°/3

x = 50°

Therefore,

∠A = x + 15° = 50° + 15° = 65°,

∠B = x – 15° = 50° – 15° = 35°,

∠C = x + 30° = 50° + 30° = 80°

6. The angles of a triangle are arranged in ascending order of their magnitude. If the difference between two consecutive angles is 10°, find the three angles.

**Solution:**

Let ∠A = x + 10°, ∠B = x + 20°, ∠C = x + 30°

Then,

∠A + ∠B + ∠C = 180° ———– (1) [from theorem 1 – interior angle theorem]

Substitute the value of ∠A, ∠B and ∠C in equation (1)

Then,

(x + 10°) + (x + 20°) + (x + 30°) = 180°

3x + 60 = 180°

3x = 120°

x = 120°/3

x = 40°

Therefore,

∠A = x + 10° = 40° + 10° = 50°,

∠B = x + 20° = 40° + 20° = 60°,

∠C = x + 30° = 40° + 30° = 70°

**3.2.3 Exterior angles**

Consider a triangle ABC. If the side BC is produced externally to form a ray BD, then ∠ACD is called an **exterior angle **of triangle ABC at C and is denoted by Ext∠C

**Theorem 2: If a side of triangle is produced, the exterior angle so formed is eequal to the sum of corresponding interior opposite angles. [ **Exterior angle theorem]

**Given: **In triangle PQR, produce QR to S. Then ∠PRS is an exterior opposite angles are ∠PRS and ∠QPR.

**To prove: **∠PRS = ∠QPR + ∠PQR

**Proof: **

**Statement:**

∠QPR + ∠PQR + ∠PRQ = 180° [Interior angle theorem]

∠PRQ + ∠PRS = 180° [linear pair]

∠QPR + ∠PQR + ∠PRQ = ∠PRQ + ∠PRS [Axiom1 – **Things which are equal to the same thing are equal to one another.**]

∠QPR + ∠PQR = ∠PRS [Axiom 2 – **If equals are added to equals, the wholes are equal.**]

This completes the proof.

**Example 4: **An exterior angle of a triangle is 100° and one of the interior opposite angles is 45°. Find the other two angles of the triangle.

**Solution:**

Let ABC be a triangle whose side BC is produced to form an exterior angle ∠ACD such that, ext∠ACD is 100°. Let ∠B = 45°.

By exterior angle theorem we have, ∠ACD = ∠B + ∠A

100° = 45° + ∠A

∠A = 100° – 45° = 55°

Hence, ∠A = 55**°**

∠C = 180° – (∠A + ∠B)

= 180° – (55**° + 45****°**)

= 80**°**

**Example 5: **In the given figure, sides QR and RQ of a triangle PQR are produced to the points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ

Solution:

Since Q, P and S all lie in the same plane,

∠QPR + ∠SPR = 180°

Hence, ∠QPR + 135° = 180°

or

∠QPR = 180° – 135° = 45°

Usinf exterior angle theorem in triangle PQR, we have

∠PQT = ∠QPR + ∠PRQ

This gives, 110° = 45° + ∠PRQ

Solving for, ∠PRQ, we get, ∠PRQ = 110° – 45° = 65°

**Example 6: **The side BC of a triangle ABC is produced on both the sides. Show that the sum of the exterior angles so formed is greater than ∠A by two right angles.

**Solution:**

Let us draw a triangle ABC, and produce BC on both sides to points D and F. Denote the angles as shown in the figure.

Now, we have to show that, ∠4 + ∠5 = ∠1 + 180°

By exterior theorem, we have,

∠4 = ∠1 + ∠3 and ∠5 = ∠1 + ∠2

Adding these two , we get,

∠4 + ∠5 = (∠1 +∠3) + (∠1 + ∠2) = ∠1 +(∠1 + ∠2 + ∠3) = ∠1 + 180°

Since the sum of all the interior angles of a triangle is 180°.

**Exercise 3.2.3**

- The exterior angles obtained on producing the base of a triangle both ways are 104° and 136°. Find the angles of the triangle.

**Solution:**

Let ABC be a given triangle with base BC. B is produced till E and C is produced till F this makes ∠ABF = 104° and ∠ACF = 136°. Now we have to find ∠ABC, ∠BCA and ∠ACB.

∠ACB + ∠ACF = 180° (linear pair)

We know,

∠ACF = 136°

Then, ∠ACB = 180° – ∠ACF = 180° – 136° = 44°

∠ABE + ∠ABC = 180° (linear pair)

We know,

∠ABE = 104°

Then, ∠ABC = 180° – ∠ABE = 180° – 104° = 76°

From interior angles theorem , sum of three interior angles is equal to 180°

Then,

∠ABC + ∠BCA + ∠CAB = 180° —————(1)

We know,

∠ABC = 76° and ∠ACB = 44°

Substitute the value of ∠ACB = 44° and ∠ABC = 76° in equation (1)

76° + 44° + ∠CAB = 180°

Then, ∠CAB = 180° – 76° – 44° = 60°

Therefore, ∠ABC = 76°

∠BCA = 44°

∠CAB = 60°

2. Sides BC, CA and AB of a triangle ABC are produced in an order, forming exterior angles ∠ACD, ∠DAE and ∠CBF. Show that ∠ACD + ∠DAE + ∠CBF = 360˚.

**Solution:**

∠ACB + ∠ACD = 180˚ ( Proposition 1: **Let AB be a straight line and OC be a ray standing on the line AB. Then ∠BOC + ∠COA = 180°)**

∠BAC + ∠CAE = 180˚ (Proposition 1: **Let AB be a straight line and OC be a ray standing on the line AB. Then ∠BOC + ∠COA = 180°**)

∠FBC + ∠ABC = 180˚ (Proposition 1: **Let AB be a straight line and OC be a ray standing on the line AB. Then ∠BOC + ∠COA = 180°**)

Adding above equations.

∠ACB + ∠ACD + ∠BAC + ∠CAE + ∠FBC + ∠ABC = 360˚ + 160˚

∠ACD + ∠CAE + ∠CBF + { ∠ACB + ∠BAC + ∠ABC } = 360˚ + 180˚

∠ACD + ∠BAE + ∠CBF + 180˚ ( Interior angle theorem ) = 360˚ + 180˚

∠ACD + ∠BAE + ∠CBF = 360˚ ( **Axiom 3: If equals are subtracted from equals, then the remainders are equal.)**

3. Compute the value of x in each of the following figures:

(i))

Solution:

Triangle ABC is an isosceles triangle. AB = AC , then ∠ABC = ∠ACB = 50°

∠ACD = x

∠ACD + ∠ACB = 180°

x = 180° – ∠ACB = 180° – 50° = 130°

(ii)

**Solution:**

∠ACB = x

∠CAD + ∠CAB = 180°

∠CAB = 180° – ∠CAD = 180° – 130° = 50°

∠ABE + ∠ABC = 180°

∠ABC = 180° – ∠ABE = 180° – 106° = 74°

From interior angles theorem, ∠ABC + ∠BCA + ∠CAB = 180°

∠BCA = 180° – 50° – 74°

= 56°

(iii)

**Solution:**

∠EAF = ∠BAC = 65° [vertically opposite angles]

∠ACD + ∠ACB = 180° [linear angle]

∠ACB = 180° – ∠ACD = 180° – 100° = 80°

From interior angle theorem, ∠BAC + ∠ACB + ∠ABC = 180°

∠ABC = 180° – ∠BAC – ∠ACB = 180° – 80° – 65° = 35°

(iv)

**Solution:**

∠ACD + ∠ACB = 180° [linear angle]

∠ACB = 180° – ∠ACD = 180° – 112° = 68°

∠EAB + ∠BAC = 180° [linear angle]

∠BAC = 180° – ∠EAB = 180° – 120° = 60°

From interior angle theorem, ∠BAC + ∠ACB + ∠ABC = 180°

∠ABC = 180° – ∠BAC – ∠ACB

x = 180° – 60° – 68°

x = 52°

(v)

**Solution:**

Triangle ABC is an isosceles triangle, AB = BC, then, ∠ACB = ∠BAC = 20°

From interior angle theorem, ∠BAC + ∠ACB + ∠ABC = 180°

∠ABC = 180° – ∠BAC – ∠ACB

∠ABC = 180° – 20° – 20°

∠ABC = 140°

∠ABD = x

∠ABD + ∠ABC = 180°

x = 180° – 140° = 40°

4. In the figure , QT ⏊ PR , TQR = 40˚ and SPR = 30˚ , Find TRS and PSQ.

**Solution:**

In Δ TRQ,

∠QTR = 90˚ ( data)

∠TQR = 40˚ (data)

∠TQR = 180˚ – ( 90˚ + 40˚ ) (Remaining angle)

= 180˚ – 130˚

∠TRQ = 50˚

In Δ PSR,

Ext ∠PSQ = ∠SPR + ∠PRS ( Exterior angle theorem)

= 30˚ + 50˚

∴Ext∠PSQ = 80˚

5. An exterior angle of triangle is 120° and one of the interior opposite angles is 30°. Find the other angles of the triangles.

**Solution:**

∠ACB + ∠ACD = 180˚ ( Proposition 1: **Let AB be a straight line and OC be a ray standing on the line AB. Then ∠BOC + ∠COA = 180°)**

∠ACB = 180˚ – 120˚

∠ACB = 60˚ ( Third angle)

In ABC,

∠A + ∠B + ∠C = 180˚

( Interior angle theorem)

∠A = 180˚- ( 30˚ + 60˚ )

∠A = 180˚ – 90˚

∠A = 90˚ ( Other interior opposite angle)

5. Find the sum of all the angles at the five vertices of the adjoining star.

Solution:

∠2 + ∠4 + ∠6 = 180˚

(Interior angle theorem)

∠1 + ∠8 + ∠10 = 180˚ ( Th .3)

∠2 + ∠9 + ∠5 = 180˚ ( Th.3)

∠1 + ∠8 + ∠4 = 180˚ ( Th.3)

∠3 + ∠7 + ∠5 = 180˚ ( Th.3)

Adding above equations,

2 ∠1 + 2 ∠2 + 2 ∠3 + 2 ∠4 + 2 ∠5 + ∠6 + ∠7 + ∠8 + ∠9 + ∠10

= 5 x 180°

2 ( ∠1 + ∠2 + ∠3 + ∠4 + ∠5 ) + ∠6 + ∠7 + ∠8 + ∠9 + ∠10 = 900°

2 ( ∠1 + ∠2 + ∠3 + ∠4 + ∠5 ) + 540˚ ( Sum of angles of a pertragon is 540˚) = 900°

2 ( ∠1 + ∠2 + ∠3 + ∠4 + ∠5 ) = 900˚ – 540˚

∠1 + ∠2 + ∠3 + ∠4 + ∠5 = 360˚/2 = 180˚

∠1 + ∠2 + ∠3 + ∠4 + ∠5 = 180˚

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