Drag and drop formal proof. Prove the Polygon Exterior Angle Sum Theorem for the enclosed triangle, that is ∠1+∠2+∠3=360° .

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°.

answer is d-29

step-by-step explanation:

since you didn't include the picture, i can give you the description so you know how to pick the correct answer. the third quartile of a data is always shown in a box plot. in the middle of your graph should be a box. the third quartile is the value that goes with the far right edge of the box.