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l M

l

l

Q____l____ N

lO

l

l

l P

Let MP & NQ be lines intersecting at O.

Then angle MON + angle NOP = 180 degrees (linear pair)

Angle MOQ + angle QOP = 180 degrees (linear pair)

Adding the equations,

Angle MON + Angle NOP + Angle MOQ + Angle QOP = 180 + 180 degrees = 360 degrees.

Therefore all angles around O sum upto 360 degrees.

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Q: Show that sum of all the angles around a point is 360?

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Because all angles around a point add up to 360 degrees.

The tessellating polygons must meet at a point. At that point, the sum of the interior angles of the polygons must 360 degrees - the sum of angles around any point. Therefore, each interior angle must divide 360 evenly. The interior angles of regular polygons with 7 or more sides lie in the range (120, 180) degrees and so cannot divide 360.

This has to do with the way in which the sum of the angles is derived. First you select a point inside the polygon and then join that point to each of the vertices. For a polygon with n sides, this gives rise to n triangles. The sum of the 3 angles of any triangle is 180 degrees. So the sum of the angles of all the triangles is n*180 degrees. Now, the "outer" angles of these triangles correspond to the interior angles of the polygon. But the sum also includes the angles formed arounf the central point. The sum of all the angles around this central point is 360 degrees. This is not part of the sum of the interior angles of the polygon and so must be subtracted. Thus, the interior angles of a polygon sum to n*180 - 360 degrees or 180*(n- 2) degrees.

The angles of a rectangle add to 360 degrees.

Angles around the circle O add up to 360 degrees Angles in Y are 2 of them are obtuse angles and 1 is an acute angle and the 3 of them add up to 360 degrees Angle in J could be a right angle

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