EXERCISE 6.3
1) Find the value of the unknown x in the following diagrams:
Ans:
i) By angle sum property of a triangle
x + 50ᵒ + 60ᵒ = 180ᵒ
x = 180ᵒ – 50ᵒ – 60ᵒ ⇒ x = 180ᵒ – 110ᵒ ⇒ x = 70ᵒ
ii) By angle sum property of a triangle
x + 30ᵒ + 90ᵒ = 180ᵒ
x + 120ᵒ = 180ᵒ ⇒x = 180ᵒ – 120ᵒ ⟹ x = 60ᵒ
iii) By angle sum property of a triangle
x + 30ᵒ + 110ᵒ = 180ᵒ
x + 140ᵒ = 180ᵒ ⟹ x = 180ᵒ – 140ᵒ ⟹ x = 40ᵒ
iv) By angle sum property of a triangle
x + x + 50ᵒ = 180ᵒ
2x + 50ᵒ = 180ᵒ ⟹ 2x = 180ᵒ – 50ᵒ ⟹ 2x = 130ᵒ ⟹ x = 130ᵒ/2
x = 65ᵒ
v) By angle sum property of a triangle
x + x + x + = 180ᵒ
3x = 180ᵒ ⟹ x = 180ᵒ/3 ⟹ x = 60ᵒ
vi) By angle sum property of a triangle
x + 2x + 90ᵒ = 180ᵒ
3x + 90ᵒ = 180ᵒ
3x = 180ᵒ – 90ᵒ ⟹ 3x = 90ᵒ ⟹ x = 90ᵒ/3 ⟹ x = 30ᵒ
2) Find the value of the unknown x and y in the following diagrams:
Ans:
i) Sum of interior opposite angles = exterior angles
x + 50ᵒ = 120ᵒ
x = 120ᵒ − 50ᵒ
x = 70ᵒ
By angle sum property of triangle
x + y + 50ᵒ = 180ᵒ
70ᵒ + y + 50ᵒ = 180ᵒ
y = 180ᵒ − 70ᵒ − 50ᵒ
y = 180ᵒ −120ᵒ ⟹ y = 60ᵒ
ii) y = 80ᵒ —– vertically opposite angles
By angle sum property of triangle
x + y + 50ᵒ = 180ᵒ
x + 80ᵒ + 50ᵒ = 180ᵒ
x = 180ᵒ − 80ᵒ − 50ᵒ
x = 180ᵒ − 130ᵒ ⟹ x = 50ᵒ
iii) Exterior angles = sum of interior opposite angles
x = 50ᵒ + 60ᵒ
x = 110ᵒ
By angle sum property of triangle
y + 50ᵒ + 60ᵒ = 180ᵒ
y + 110ᵒ = 180ᵒ
y = 180ᵒ − 110ᵒ
y = 70ᵒ
iv) x = 60ᵒ —- vertically opposite angles
By angle sum property of triangle
x + y + 30ᵒ = 180ᵒ
60ᵒ + y + 30ᵒ = 180ᵒ
y = 180ᵒ − 60ᵒ – 30ᵒ
y = 180ᵒ − 90ᵒ ⟹ y = 90ᵒ
v) y = 90ᵒ —- vertically opposite angles
By angle sum property of triangle
x + x + y = 180ᵒ
2 x + 90ᵒ = 180ᵒ
2 x = 180ᵒ − 90ᵒ
2 x = 90ᵒ ⟹ x = 90ᵒ/2 ⟹ x = 45ᵒ
vi) From the rule of vertically opposite angles
x = y
we know, sum of all interior angles of a triangle is 180ᵒ
x + x + x = 180ᵒ
3 x = 180ᵒ ⟹ x = 180ᵒ/3 ⟹x = 60ᵒ
Since x = y
∴ y = 60ᵒ