In this chapter, we provide KSEEB SSLC Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.3 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.3 pdf, free KSEEB SSLC Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.3 pdf download. Now you will get step by step solution to each question.

**Karnataka State Syllabus Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.3**

## Karnataka Board Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.3

Question 1.

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

Answer:

In triangle ABC, BC is produced on either sides.

Let ∠ABC = 140° and ∠ACE = 136°

∠ABD +∠ABC = 180° [ Linear pair]

104 + ∠ABC = 180°

∠ABC = 180 – 104

∠ ABC = 76°

∠ACB + ∠ACE = 180° [Linearpair]

∠ACB + 136° = 180°

∠ACB = 180 – 136 ∠ACB = 44°

∠ABD + ∠ACB + ∠BAC = 180°

[Sum of the angles of a triangle 180° ]

76 + 44 + ∠BAC = 180°

120 + ∠BAC = 180°

∠BAC = 180 – 120 ∠BAC = 60°

Three angles are 76°, 44° & 60°

Question 2.

Sides BC, CA and AB of a triangle ABC are produced in order, forming exterior angles ∠ACD, ∠BAE, and∠CBF.

Show that ∠ACD + ∠BAE + ∠CBF = 360°.

Answer:

ABC + ∠CBF = 180° [Linear pair]………..(i)

∠ACB + ∠ACD = 180° [Linear pair]……(ii)

∠BAC + ∠BAE = 180° [Linear pair]…….(iii)

Byadding(i), (ii) and (iii)

∠ABC + ∠CBF + ∠ACB + ∠ACD+

∠BAC + ∠BAE = 180 + 180 + 180

∠ABC + ∠ACB + ∠BAC +

∠CBF + ∠ACD + ∠BAE = 540°

180 + ∠CBF + ∠ACD + ∠BAE = 540°

[Sum of the angles of a triangle 180°]

∠CBF + ∠ACD + ∠BAE = 360°

Question 3

Compute the value of x in each of the following figures

Answer:

i.

AB = AC

∴ ∠ABC = ∠ACB = 50°

∠ACB + ∠ACD = 180° [Linear pair]

50 + x = 180°

x = 180 – 50 = 130°

ii.

∠ABC + ∠CBF = 180° [Linear pair].

106°+∠ABC = 180°

∠ABC = 180 – 106 = 74°

∠EAC = ∠ABC + ∠ACB

130° =74 +x

130 – 74 = x

∴x = 56°

iii.

∠QPR = ∠TPU = 65°

[Vertically opposite angles]

∠PRS = ∠PQR + ∠QPR [Exterior angle=Sum of interior opposire angle]

100 = 65 + x

100 – 65 = x

35 = x

∴ x = 35°.

iv.

∠BAE + ∠BAC = 180° [Linear pair]

120° + ∠BAC = 180°

∠BAC = 180 – 120

∠BAC = 60°

∠ACD = ∠BAC + ∠ABC

[Exterior angle = Sum of interior opposuite angle]

112° = 60+x

112 – 60 = x

52 = x

x = 52°

v.

In Δ ABC, BA = BC (data)

∴ ∠BAC = ∠BCA = 20°

[Base angles of an isosceles traingle]

∠ABD = ∠BAC + ∠BCA

[Exterior angle=Sum of interior opposite angle]

x = 20 + 20

x=40°

Question 4.

In figure QT ⊥ PR, ∠TOR = 40° and ∠SPR = 30° find ∠TRS and ∠PSQ.

Answer:

In ΔTQR,

∠TQR + ∠QTR + ∠TRQ = 180°

40 + 90 + ∠TRQ = 180°

130 + ∠TRQ = 180

∠ TRQ = 180 – 130

∠ TRQ = 50°

∠ TRS = 50° [∠PRS is same as ∠TRS ]

In Δ PRS, RS is produced to Q

∴Exterior∠PSQ = ∠SPR + ∠PRS = 30 + 50

∠PSQ = 80°

Question 5.

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

Answer:

In Δ ABC, BC is produced to D. Let

∠ACD = 120° and ∠ABC = 30°

Exterior ∠ACD = ∠BAC + ∠ABC

120 = ∠BAC + 30

120 – 30 = ∠BAC

90° = ∠BAC

∴∠BAC = 90

∠ACB +∠ACD = 180° [Linear pair]

∠ACB + 120 = 180°

∠ACB = 180 – 120

∠ACB = 60°

Other two angles are 90° & 60°

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