In this chapter, we provide KSEEB SSLC Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.2 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 6.2 pdf, free KSEEB SSLC Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.2 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 6.2**

## Karnataka Board 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°

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 = 1302

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 = 1806 = 30

x = 30°

2x = 2 × 30 = 60°

3x = 3 × 30 = 90°

∴ The angles are 30°, 60° and 90°

Question 5.

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

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 = 1503 = 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°

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 = 1503 = 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°

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