NCERT Solutions Class 7th Math Chapter – 6 The Triangle and its Properties Exercise – 6.2

(i)

Solution: We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x = 50^o + 70^o
= x = 120^o

Solution: We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x = 65^o + 45^o
= x = 110^o

Solution: We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x = 30^o + 40^o
= x = 70^o

Solution:  We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x = 60^o + 60^o
= x = 120^o

Solution:  We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x = 50^o + 50^o
= x = 100^o

Solution: We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x = 30^o + 60^o
= x = 90^o

(i)

Solution: We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x + 50^o = 115^o
By transposing 50o from LHS to RHS it becomes – 50^o
= x = 115^o – 50^o
= x = 65^o

Solution:  We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= 70^o + x = 100^o
By transposing 70^o from LHS to RHS it becomes – 70^o
= x = 100^o – 70^o
= x = 30^o

Solution: We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
The given triangle is a right-angled triangle. So the angle opposite to the x is 90^o.
= x + 90^o = 125^o
By transposing 90^o from LHS to RHS it becomes – 90^o
= x = 125^o – 90^o
= x = 35^o

Solution: We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x + 60^o = 120^o
By transposing 60o from LHS to RHS it becomes – 60^o
= x = 120^o – 60^o
= x = 60^o

Solution: We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
The given triangle is a right-angled triangle. So the angle opposite to the x is 90^o.
= x + 30^o = 80^o
By transposing 30o from LHS to RHS it becomes – 30^o
= x = 80^o – 30^o
= x = 50^o

Solution: We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
The given triangle is a right-angled triangle. So the angle opposite to the x is 90^o.
= x + 35^o = 75^o
By transposing 35o from LHS to RHS it becomes – 35^o
= x = 75^o – 35^o
= x = 40^o