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