NCERT Solutions Class 7th Math Chapter – 6 The Triangle and its Properties Exercise 6.3

(i)

Solution: We know that,
The sum of all the interior angles of a triangle is 180^o.
Then,
= ∠BAC + ∠ABC + ∠BCA = 180^o
= x + 50^o + 60^o = 180^o
= x + 110^o = 180^o
By transposing 110o from LHS to RHS it becomes – 110^o
= x = 180^o – 110^o
= x = 70^o

Solution:  We know that,
The sum of all the interior angles of a triangle is 180^o.
The given triangle is a right angled triangle. So the ∠QPR is 90^o.
Then,
= ∠QPR + ∠PQR + ∠PRQ = 180^o
= 90^o + 30^o + x = 180^o
= 120^o + x = 180^o
By transposing 110o from LHS to RHS it becomes – 110^o
= x = 180^o – 120^o
= x = 60^o

Solution:  We know that,
The sum of all the interior angles of a triangle is 180^o.
Then,
= ∠XYZ + ∠YXZ + ∠XZY = 180^o
= 110^o + 30^o + x = 180^o
= 140^o + x = 180^o
By transposing 140o from LHS to RHS it becomes – 140^o
= x = 180^o – 140^o
= x = 40^o

Solution: We know that,
The sum of all the interior angles of a triangle is 180^o.
Then,
= 50^o + x + x = 180^o
= 50^o + 2x = 180^o
By transposing 50o from LHS to RHS it becomes – 50^o
= 2x = 180^o – 50^o
= 2x = 130^o
= x = 130^o/2
= x = 65^o

Solution:  We know that,
The sum of all the interior angles of a triangle is 180o.
Then,
= x + x + x = 180^o
= 3x = 180^o
= x = 180^o/3
= x = 60^o
∴The given triangle is an equiangular triangle.

Solution:  We know that,
The sum of all the interior angles of a triangle is 180^o.
Then,
= 90^o + 2x + x = 180^o
= 90^o + 3x = 180^o
By transposing 90^o from LHS to RHS it becomes – 90^o
= 3x = 180^o – 90^o
= 3x = 90^o
= x = 90^o/3
= x = 30^o
Then,
= 2x = 2 × 30^o = 60^o

(i)

Solution: We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
Then,
= 50^o + x = 120^o
By transposing 50^o from LHS to RHS it becomes – 50^o
= x = 120^o – 50^o
= x = 70^o 
We also know that,
The sum of all the interior angles of a triangle is 180^o.
Then,
= 50^o + x + y = 180^o
= 50^o + 70^o + y = 180^o
= 120^o + y = 180^o
By transposing 120^o from LHS to RHS it becomes – 120^o
= y = 180^o – 120^o 
= y = 60^o

Solution: From the rule of vertically opposite angles,
= y = 80^o
Then,
We know that,
The sum of all the interior angles of a triangle is 180^o.
Then,
= 50^o + 80^o + x = 180^o
= 130^o + x = 180^o
By transposing 130^o from LHS to RHS it becomes – 130^o
= x = 180^o – 130^o 
= x = 50^o

Solution:  We know that,
The sum of all the interior angles of a triangle is 180o.
Then,
= 50^o + 60^o + y = 180^o
= 110^o + y = 180^o
By transposing 110^o from LHS to RHS it becomes – 110^o
= y = 180^o – 110^o 
= y = 70^o
Now,
From the rule of linear pair,
= x + y = 180^o
= x + 70^o = 180^o
By transposing 70^o from LHS to RHS it becomes – 70^o
= x = 180^o – 70^o 
= x = 110^o

Solution: From the rule of vertically opposite angles,
= x = 60^o
Then,
We know that,
The sum of all the interior angles of a triangle is 180^o.
Then,
= 30^o + x + y = 180^o
= 30^o + 60^o + y = 180^o
= 90^o + y = 180^o
By transposing 90^o from LHS to RHS it becomes – 90^o
= y = 180^o – 90^o 
= y = 90^o

Solution:  From the rule of vertically opposite angles,
= y = 90^o
Then,
We know that,
The sum of all the interior angles of a triangle is 180^o.
Then,
= x + x + y = 180^o
= 2x + 90^o = 180^o
By transposing 90o from LHS to RHS it becomes – 90^o
= 2x = 180^o – 90^o 
= 2x = 90^o
= x = 90^o/2
= x = 45^o

Solution:  From the rule of vertically opposite angles,
= x = y
Then,
We know that,
The sum of all the interior angles of a triangle is 180^o.
Then,
= x + x + x = 180^o
= 3x = 180^o
= x = 180^o/3
= x = 60^o