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.
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.
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Theoretical proof:
Given: 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.
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.
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.
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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