Theorems onTriangles

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 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:

VerticesSidesAngles
AAB∠BAC or ∠A
BBC∠ABC or ∠B
CCA∠BCA or ∠C

0.1

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:

1

Classification based on angles of the triangle:

There 3 types of triangle:

(i) Acute angled triangle

(ii) Right angled triangle

(iii) Obtuse angled triangle

2

Exercise 3.2.1

1. Match the following:

(i)

3                                           (a) Equilateral triangle

(ii)

4                                     (b) Acute angled triangle

(iii)

5                                               (c) Right angled triangle

(iv)

6                                               (d) Obtuse angled triangle

Solution:

3   ————— Right angled triangle

4—————– Obtuse angled triangle

5——————- equilateral triangle

6——————— Acute angled triangle

2. Based the sides, classify the following triangles:

(i)

7.1

Solution:

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

(ii)

7.2

Solution:

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

(iii)

7.3

Solution:

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

(iv)

7.4

Solution:

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

(v)

7.5

Solution:

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

(vi)

7.6

Solution:

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

(vii)

7.7

Solution:

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

(viii)

7.8

Solution:

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

(ix)

7.9

Solution:

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

(x)

7.10

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

theorem1

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:

e1

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:

e2

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.

e3

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

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

Solution:

ex1

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°

 

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

Solution:

ex2

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:

ex3

 

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:

ex4

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.

ex45

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:

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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

ex2.4

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.

teorem 2

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°.

ex2.4

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:

example 5

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.

example 6

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

  1. 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.

262.1

∠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:

 

hhhhhhhhhhhhhhhhhhhhh

∠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))

262.3.1

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)

262.3.2

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)

262.3.3

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)

262.3.6

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)

262.3.5

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.

262.4

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:

262.5

∠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.

262. 6

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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