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

**NOTE: We know that the sum of all the interior angles of a triangle is 180 ^{o}.**

(i)

**Solution:**

∠BAC + ∠ABC + ∠BCA = 180^{o}

x + 50^{o} + 60^{o} = 180^{o}

x + 110^{o} = 180^{o}

Transposing 110^{o} from LHS to RHS it becomes – 110^{o}

x = 180^{o} – 110^{o}

x = 70^{o}

(ii)

**Solution:**

The given triangle is a right-angled triangle. So the ∠QPR is 90^{o}.

Now,

∠QPR + ∠PQR + ∠PRQ = 180^{o}

90^{o} + 30^{o} + x = 180^{o}

120^{o} + x = 180^{o}

Transposing 110^{o} from LHS to RHS it becomes – 110^{o}

x = 180^{o} – 120^{o}

x = 60^{o}

(ii)

**Solution:**

∠XYZ + ∠YXZ + ∠XZY = 180^{o}

110^{o} + 30^{o} + x = 180^{o}

140^{o} + x = 180^{o}

Transposing 140^{o} from LHS to RHS it becomes – 140^{o}

x = 180^{o} – 140^{o}

x = 40^{o}

**Solution:**

50^{o} + x + x = 180^{o}

50^{o} + 2x = 180^{o}

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:**

x + x + x = 180^{o}

3x = 180^{o}

x = 180^{o}/3

x = 60^{o}

Hence, the given triangle is an equilateral triangle.

(vi)

**Solution:**

90^{o} + 2x + x = 180^{o}

90^{o} + 3x = 180^{o}

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:

**Solution:**

(i)

**Solution: **

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

Now,

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}

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

Now,

50^{o} + x + y = 180^{o}

50^{o} + 70^{o} + y = 180^{o}

120^{o} + y = 180^{o}

Transposing 120^{o} from LHS to RHS it becomes – 120^{o}

y = 180^{o} – 120^{o}

y = 60^{o}

(ii)

**Solution:**

According to the rule of vertically opposite angles,

y = 80^{o}

Now,

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

50^{o} + 80^{o} + x = 180^{o}

130^{o} + x = 180^{o}

Transposing 130^{o} from LHS to RHS it becomes – 130^{o}

x = 180^{o} – 130^{o}

x = 50^{o}

(iii)

**Solution:**

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

50^{o} + 60^{o} + y = 180^{o}

110^{o} + y = 180^{o}

Transposing 110^{o} from LHS to RHS it becomes – 110^{o}

y = 180^{o} – 110^{o}

y = 70^{o}

Now,

According to the rule of linear pair,

x + y = 180^{o}

x + 70^{o} = 180^{o}

Transposing 70^{o} from LHS to RHS it becomes – 70^{o}

x = 180^{o} – 70^{o}

x = 110^{o}

(iv)

**Solution:**

According the rule of vertically opposite angles,

x = 60^{o}

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

30^{o} + x + y = 180^{o}

30^{o} + 60^{o} + y = 180^{o}

90^{o} + y = 180^{o}

Transposing 90^{o} from LHS to RHS it becomes – 90^{o}

y = 180^{o} – 90^{o}

y = 90^{o}

(v)

**Solution:**

According to the rule of vertically opposite angles,

y = 90^{o}

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

Now,

x + x + y = 180^{o}

2x + 90^{o} = 180^{o}

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:**

According to the rule of vertically opposite angles,

x = y

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

Now,

x + x + x = 180^{o}

3x = 180^{o}

x = 180^{o}/3

x = 60^{o}

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