**1. Find the value of the unknown exterior angle x in the following diagram:**

**Solution:**

**(i)**

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}

**(ii)**

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}

**(iii)**

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}

**(iv)**

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}

**(v)**

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}

**(vi)**

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}

**2. Find the value of the unknown interior angle x in the following figures:**

**Solution:**

**(i)**

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.

x + 50^{o} = 115^{o}

Transposing (50^{o}) from LHS to RHS then it becomes (– 50^{o})

x = 115^{o} – 50^{o}

x = 65^{o}

**(ii)**

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.

70^{o} + x = 100^{o}

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

x = 100^{o} – 70^{o}

x = 30^{o}

**(iii)**

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.

Here the given triangle is a right-angled triangle. So the angle opposite to the x is 90^{o}.

x + 90^{o} = 125^{o}

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

x = 125^{o} – 90^{o}

x = 35^{o}

(**iv)**

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 (60^{o}) from LHS to RHS it becomes (– 60^{o})

x = 120^{o} – 60^{o}

x = 60^{o}

**(v)**

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.

Here the given triangle is a right-angled triangle. So, the angle opposite to the x is 90^{o}.

x + 30^{o} = 80^{o}

Transposing (30^{o}) from LHS to RHS it becomes (– 30^{o})

x = 80^{o} – 30^{o}

x = 50^{o}

**(vi)**

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.

Here 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 (35^{o})from LHS to RHS it becomes (– 35^{o})

x = 75^{o} – 35^{o}

x = 40^{o}

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