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 = 50o + 70o
x = 120o
(ii)

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.
x = 65o + 45o
x = 110o
(iii)

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.
x = 30o + 40o
x = 70o
(iv)

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.
x = 60o + 60o
x = 120o
(v)

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.
x = 50o + 50o
x = 100o
(vi)

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.
x = 30o + 60o
x = 90o
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 + 50o = 115o
Transposing (50o) from LHS to RHS then it becomes (– 50o)
x = 115o – 50o
x = 65o
(ii)

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.
70o + x = 100o
Transposing (70o) from LHS to RHS it becomes (– 70o)
x = 100o – 70o
x = 30o
(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 90o.
x + 90o = 125o
Transposing (90o) from LHS to RHS it becomes (– 90o)
x = 125o – 90o
x = 35o
(iv)

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.
x + 60o = 120o
By transposing (60o) from LHS to RHS it becomes (– 60o)
x = 120o – 60o
x = 60o
(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 90o.
x + 30o = 80o
Transposing (30o) from LHS to RHS it becomes (– 30o)
x = 80o – 30o
x = 50o
(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 90o.
x + 35o = 75o
By transposing (35o)from LHS to RHS it becomes (– 35o)
x = 75o – 35o
x = 40o
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