**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}