1. Find the value of the unknown x in the following diagrams:

(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 110^o from LHS to RHS it becomes – 110^o

= x = 180^o – 110^o

= x = 70^o

(ii)

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 110^o from LHS to RHS it becomes – 110^o

= x = 180^o – 120^o

= x = 60^o

(iii)

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 140^o from LHS to RHS it becomes – 140^o

= x = 180^o – 140^o

= x = 40^o

(iv)

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 50^o from LHS to RHS it becomes – 50^o

= 2x = 180^o – 50^o

= 2x = 130^o

= x = 130^o/2

= x = 65^o

(v)

Solution:-

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

∴The given triangle is an equiangular triangle.

(vi)

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

2. Find the values of the unknowns x and y in the following diagrams:

(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

(ii)

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

(iii)

Solution:-

We know that,

The sum of all the interior angles of a triangle is 180^o.

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

= x = 110^o

(iv)

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

(v)

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 90^o from LHS to RHS it becomes – 90^o

= 2x = 180^o – 90^o 

= 2x = 90^o

= x = 90^o/2

= x = 45^o

(vi)

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