KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.2

Students can Download Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

Karnataka State Syllabus Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.2

Question 1.
In a triangle ABC, if ∠A= 55° and∠B.= 40° find ∠C
Answer:
∠A +∠B + ∠C = 180°
[Sum of the angles of a triangle 180°]
55 + 40 + ∠C = 180°
95 + ∠C = 180“
∠C = 180 – 95
∠C = 85°

Question 2.
In a right-angled triangle, if one of the other two angles is 35°, find the remaining angle.
Answer:
Let the angles be ∠A, ∠B and ∠C
Then ∠A = 90°,∠B = 35° and∠C = ?
∠A + ∠B + ∠C = 180°
90 + 35 + ∠C = 180°
125 + ∠C = 180
∠C = 180 – 125 = 55°
∠C = 55°

KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.2

Question 3.
If the vertex angle of an isosceles triangle is 50° find the other two angles.
Answer:
In an isosceles triangle, the base angles are equal, Let the each base angle be ‘x’
∠A +∠B +∠C = l80°
50 + x + x = 180°
[Sum of the angles of a traingle 180° ]
50 + 2x = 180
2x = 180 – 50
2x = 130
x = \(\frac { 130 }{ 2 }\)
x = 65°
∴The other two angles are equal to 65° & 65°

Question 4.
The angles of a triangle are in the ratio 1 : 2 : 3. Determine the three angles.
Answer:
Let the common ratio be ‘x’
The three angles are x,2x and 3x
x + 2x + 3x = 180°
[Sum of the angles of a triangle 180°]
6x = 180°
x = \(\frac { 180 }{ 6 }\) = 30
x = 30°
2x = 2 × 30 = 60°
3x = 3 × 30 = 90°
∴ The angles are 30°, 60° and 90°

KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.2

Question 5.
In the adjacent triangle ABC, find the value of x and calculate the measure of all the angles of the triangle.
KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.2 1
Answer:
∠A + ∠B + ∠C = 180°
[Sum of the angles of a triangle 180° ]
x + 15 + x – 15 + x + 30 = 180°
3x + 30 = 180°
3x = 180° – 30
3x = 150
x = \(\frac { 150 }{ 3 }\) = 50°
∠A = x +15 = 50 + 15 = 65°
∠B = x – 15 = 50 – 15 = 35°
∠C = x + 30 = 50 + 30 = 80°
∴ The angles are 65°, 3 5° and 80°

KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.2

Question 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.
Answer:
Let the first angle be x, second angle is x +10° and third angle is x + 20 x + x+ 10 + x + 20 = 180°
[Sum of the angles of a triangle 180° ]
3x + 30 = 180°
3x = 180 – 30
3x = 150°
x = \(\frac { 150 }{ 3 }\) = 50°
x=50°
∴ First angle x = 50°
Second angle = x + 10 = 50 + 10 = 60°
Third angle = x + 20 = 50 + 20 = 70°
Three angles are 50°, 60° & 70°

KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.2