 Hello and welcome to the session. Let's work out the following problem. It says prove that the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle using the above find the value of x in this figure. Let's now move on to the solution and let's first prove the first part of the given question and let's first write what is given to us. We are given a circle with center O and arc AB subtending angle AOB at the center angle ACB at any point C on the remaining part of the circle. So AB is the arc subtending an angle AOB at the center and angle ACB at any point on the remaining part of the circle. Let's now write what we have to prove. We have to prove that the angle subtended by the arc at the center is double the angle subtended by the arc at any point of the remaining part of the circle. So we have to prove that angle AOB is twice the angle ACB. Let's now do some construction. Join CO produce it to a point P. Also join OA and OV that we have already joined. Let's now start the prove. In triangle AOC, OA is equal to OC because OA and OC both have the radius of the circle. This is radii of circle and then angle OCA OCA is equal to angle OAC that is this angle is equal to this angle as we know that angles opposite the equal sides are equal. These are angles opposite equal sides. Also angle POA is equal to angle OCA plus angle OAC that is this angle is the sum of these two angles right and this is because this is the exterior angle of the triangle. Therefore, angle POA is equal to angle OCA plus angle OCA since angle OCA is equal to angle OAC. So we have angle POA is equal to angle OCA plus angle OCA. So we have angle POA is equal to twice of angle OCA. Similarly, by taking triangle BOC will obtain the fact that angle POB that is this angle will be equal to angle twice of angle OCB. Let's name this as one and this as two. Now adding one and two we have angle POA plus angle POB is equal to twice of angle OCA plus twice of angle OCB that is twice of angle OCB plus angle OCB but angle OCA plus angle OCB is this complete angle and angle POA plus POB is this complete angle which is AOB therefore angle AOB is equal to twice of angle ACB because angle OCA plus angle OCB is angle ACB and that is what we have to prove that angle subtended by the arc at the center is double the angle subtended by this arc at any point of the remaining part of the circle. So hence proved. Now in the second part we have to find the value of X now we have angle AOB is equal to 40 degrees and angle ACB is X and we have to find the value of X now from the above result angle AOB is twice the angle ACB right now AOB is 40 degrees is equal to twice of angle ACB which is X so this implies X is 40 degrees by 2 that is 20 degrees hence the value of X is 20 degrees so this completes the question and the session by for now take care have a good day.