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Exterior Angles of a Triangle

If the side (s) of a triangle is produced to form a ray, then the angle formed is called exterior angle of the triangle.

Example : When side BC of Δ ABC is produced to form ray BD, then ACD is called exterior angle of Δ ABC at C and is denoted by ext. ACD. With respect to ext. ACD of Δ ABC at C, the angles A and B are called remote interior angles or interior opposite angles.


Now, if we produce AC to form a ray AE, then BCE will be the exterior angle of Δ ABC at C. Thus at each vertex of a triangle, there are two exterior angles of the triangle and these angles will be equal because they are vertically opposite angles.


Theorem 9
If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.
Given: In Δ ABC, BC is produced to form an angle ACD,
To Prove: ACD = BAC + ABC i.e. 4 = 1 + 2

Proof: In Δ ABC, we have
1 + 2 + 3 = 180° (i)
Also 3 + 4 = 180° (ii) (linear pair)
From equation (i) and equation (ii), we get
1 + 2 + 3 = 3 + 4
4 = 1 + 2

[Note: Since 4 is the sum of 1 and 2, it is clear that 4 > 2 and 4 > 1. Such a result, which is an easy consequence of a theorem, is called corollary of that theorem. Now we may state that:



An exterior angle of a triangle is greater than either of the interior opposite angles. This is also called as the Exterior Angle Theorem.



An exterior angle of a triangle is 120°, and one of the interior opposite angles is 35°. Find the other two angles of the triangle.


Let ABC be a triangle whose BC side is produced to form exterior ACD.
ACD = 120° and ABC = 35°


Now by exterior angle theorem, we have
120° = 35° + BAC
BAC = 120° - 35° = 85°
Now in Δ ABC, we have
A + B + ACB = 180°
85° + 35° + ACB = 180°
ACB = 180° - 85° - 35° = 60°
Hence two other angles are 85° and 60°.

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