Exercise 6.3
QUESTION 1
Find the value of the unknown x in the following diagrams :
The sum of all interior angles of a triangle is 180°. By using this property, these problems can be
solved as follows.
(i) x + 50° + 60° = 180°
x + 110° = 180°
x = 180° -110°
x = 70°
(ii) x + 90° + 30° = 180°
x + 120° = 180°
x = 180° – 120°
x = 60°
(iii) x + 30° + 110° = 180°
x + 140° = 180°
x = 180° – 140°
x = 40°
(iv) 50° + x + x = 180°
50° + 2x = 180°
2x = 180° – 50°
(v) x + x + x = 180°
3x = 180° – 90°
QUESTION 2
Find the value of the unknowns x and y in the following diagrams:
Sol :
(i) y+ 120° = 180° (Linear pair)
y = 180° – 120°
y = 60°
x + y + 50° = 180° (Angle sum property)
x + 60° + 50° = 180°
x + 110° = 180°
x = 180° – 10°
x = 70°
(ii) y = 80° (Vertically opposite angles)
y + x + 50° = 180° (Angle sum property)
80° + x + 50° =180°
x + 130° = 180°
x = 180° – 130°
x = 50°
(iii) y + 50° + 60° = 180° (Angle sum property)
y = 180° – 60° – 50°
y = 70°
x + y = 180° (Linear pair)
x = 180° – y
x = 180°-70°
x = 110°
(iv) x = 60° (Vertically opposite angles)
30° + x + y = 180°
30° + 60° + y = 180°
y = 180° – 30° – 60°
y = 90°
(v) y = 90° (Vertically opposite angles)
x + x + y = 180° (Angle sum property)
2x + y = 180°
2x + 90° = 180°
2x = 180° – 90°
(vi)
y = x ( Vertically opposite angles )
a = x ( Vertically opposite angles )
b = x ( Vertically opposite angles )
a + b+ y=180° (Angle sum property)
x + x + x = 180°
3x = 180°
y = x = 60°
