Answer:
The sum of all the external angles of a triangle is 360°.
Step-by-step explanation:
Let there are n sides in a polygon.
So, there are n internal angles in the polygon, let ∠i1, ∠i2, ∠i3, ..., ∠in are the measure of n internal angles of the polygon.
The measure of external angle corresponding to ∠i1, ∠1= 180°-∠i1,
The measure of external angle corresponding to ∠i2, ∠2 = 180°-∠i2,
Similarly, the measure of external angle corresponding to ∠in, ∠n = 180°-∠in.
Now, the sum of all the external angles of the polygon,
(∠1+∠2+...+∠n)=(180°-∠i1)+(180°-∠i2)+...+(180°-∠in)
=(180°+180°+...n times)-(∠i1+∠i2+...+∠in)
=n x 180° - (∠i1+∠i2+...+∠in)
As ∠i1+∠i2+...+∠in is the sum of all the internal angles of the polygon.
So, the sum of all the external angles of the polygon =
(n x 180°) - (sum of all the internal angles of the polygon).
In the case of a triangle, n=3 and the sum of all the three internal angles, ∠i1+∠i2+∠i3 = 180°.
Therefore, the sum of all the external angles of a triangle,
∠1+∠2+∠3 =3x180°-(∠i1+∠i2+∠i3)
=540°=180°
=360°.
Hence, the sum of all the external angles of a triangle is 360°.
