krupakar
1. The sum of all angles in a triangle = 180° Therefore 105° + 30° + 3rd angle = 180° 3rd angle = 180° - 135° = 45° 2. Let the other two equal angles = x° Sum of all angles in a triangle = 180° 130° + x° + x° = 180° 2 x° = 50° ⇒ x = 25° 3. Ratio of three angles = 1:2:3 Let the three angles are = 1 x°, 2 x°, 3 x° sum of all angles in a triangle = 180° 1 x + 2 x + 3 x = 180 6 x = 180 ⇒ x = 30° The three angles are 30°,60°,90° 4. 1. Exterior angle 110° = sum of 2 opposite interior angles 50 + x° x =110 - 50 = 60° 2. x = 50°(vertically opposite angles are equal) 50 + y + 70 = 180 ⇒ y = 60° 3. 90 + x + x = 180 ⇒ 2 x = 90 ⇒ x = 45° 4. exterior angle y = 45+35 = 80° x + 80° = 180° (since straight line) ⇒ x = 100° 5. x+x+x = 180° 3 x = 180° ⇒ x = 60° 6. y = 45°(vertically opposite angles are equal) x + 45° = 180° x = 135°
Answers & Comments
Therefore 105° + 30° + 3rd angle = 180°
3rd angle = 180° - 135° = 45°
2. Let the other two equal angles = x°
Sum of all angles in a triangle = 180°
130° + x° + x° = 180°
2 x° = 50°
⇒ x = 25°
3. Ratio of three angles = 1:2:3
Let the three angles are = 1 x°, 2 x°, 3 x°
sum of all angles in a triangle = 180°
1 x + 2 x + 3 x = 180
6 x = 180 ⇒ x = 30°
The three angles are 30°,60°,90°
4. 1. Exterior angle 110° = sum of 2 opposite interior angles 50 + x°
x =110 - 50 = 60°
2. x = 50°(vertically opposite angles are equal)
50 + y + 70 = 180
⇒ y = 60°
3. 90 + x + x = 180
⇒ 2 x = 90 ⇒ x = 45°
4. exterior angle y = 45+35 = 80°
x + 80° = 180° (since straight line)
⇒ x = 100°
5. x+x+x = 180°
3 x = 180° ⇒ x = 60°
6. y = 45°(vertically opposite angles are equal)
x + 45° = 180°
x = 135°