Exterior Angle of a Triangle

Exterior Angle of a Triangle

Exterior Angle of a Triangle

When a side of a triangle is produced, the angle formed outside of the triangle is called an exterior angle of the triangle. In the given figure, ABC is a triangle and ACD is an exterior angle.
exterior angle ∠ACD
The exterior angle of a triangle is always equal to the sum of two opposite interior angles of the triangle. In the above figure, exterior angle ACD is equal to the sum of two opposite interior angles ABC and BAC i.e. ACD = ABC + BAC.

Theoretical proof:

Given: ABC is a triangle with exterior angle ACD.
ABC is a triangle with exterior angle ∠ACD
To prove: ABC + BAC = ACD

Proof:
              Statements                                                    Reasons
1.    ABC + BAC + ACB = 180°----------> sum of angles of a Δ.
2.    ACB + ACD = 180° ---------> linear pair of angles.
3.    ABC + BAC + ACB = ACB + ACD -----> from statements 1 and 2.
4.    ABC + BAC = ACD -----> removing ACB from both sides .
                                                                                                            Proved.

Workout Examples

Example 1: Find the values of x from the given figure.
Example 1: Triangle ABC

Solution: From the figure,
                          x + 50° = 110° ----> exterior angle of a Δ is equal to the sum of opposite interior angles.
                or,     x = 110° – 50°
                            = 60°


Example 2: Find the values of x and y in the given figure.
Example 2: Triangle ABC

Solution: From the figure,
                          x = 35° + 75° ----> exterior angle of a Δ is equal to the sum of opposite interior angles.
                or,     x = 110°

                          x + y = 180° ----------> linear pair of angles.
                or,     110° + y = 180°
                or,     y = 180° – 110° 
                or,     y = 70°

                 x = 110° and y = 70°


Example 3: Find the values of a and b in the given figure.
Example 3: Triangle ABC

Solution: From the figure,
                          a = 50° ------> base angles of an isosceles Δ.

                          b = a + 50° ------> exterior angle of a Δ is equal to the sum of opposite interior angles.
                             = 50° + 50°
                             = 100°

                 a = 50° and b = 100°


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