 Hello and welcome to the session. In this session we discuss the final question which says if each imperial angle of a regular polygon is m times as large as each exterior angle, prove that the number of sides is 2 into m plus 1. Before we move on to the solution, let's discuss few results. First we have each imperial angle of a regular polygon is equal to 2n minus 4 upon n into right angles. Then next we have regular polygon m sides is equal to 4 upon n into the idea that we use for this question. If we see it with the solution now, according to the question we have that each interior angle of a regular polygon is m times as large as the exterior angle and we are supposed to prove that the number of sides is 2 into m plus 1. Each interior angle a regular polygon is equal to exterior angle of a regular polygon. Key idea we know that each interior angle of a regular polygon of m sides is given by 2n minus 4 upon n into right angles. So here we have minus 4 upon n into 90 degrees since each right angle is of measure 90 degrees and this is equal to m into each exterior angle of a regular polygon and this is equal to 4 upon n into right angles. Now here this 90 degrees, 90 degrees cancels, m cancels with n and we have 2n minus 4 is equal to 4n for the 2n is equal to 4n plus and from here we have 2n is equal to 4 into m plus 1 the whole now 2 times is 4 therefore we have n is equal to 2 into n plus 1 and we know that m is the number of sides of the regular polygon therefore number of sides of the regular polygon is equal to 2 into n plus 1 the whole and this is what we were supposed to throw, we have thrown that the number of sides of the regular polygon is 2 into n plus 1 the whole so this can be C session, hope you have understood the solution of this question.