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

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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