 Hello and welcome to the session. The given question says, if a line touches a circle and from the point of contact a chord is drawn, prove that the angles which the squad make with the given line are equal respectively to the angles formed in the corresponding alternate segments. By using the above, do the following, n-figure 4 which is this SPT is a tangent to the circle at P and PQ is a chord of the circle. If angle PRQ is 40 degrees and angle PQR is 80 degrees, find X and Y. First let us solve the theorem which is theorem 10.3 of your NC hearted textbook. It says if a line touches a circle and from the point of contact here, the point of contact is A and a tangent PQ touches the circle at A. If a chord is drawn, suppose AB is a chord, the angles which this chord make with the given line are equal respectively to the angles formed in the corresponding alternate segments. That is, we have to show that angle BAQ is equal to angle BCA and angle BAP is equal to angle BDA. So first let us write down what we are given. Here we are given that PQ is a tangent to a circle with center O at a point A and AB is a chord C and D are points in two segments of the circle formed by the chord AB. This figure shows this and we have to prove first that angle BAQ is equal to angle ACB and then we have to prove that angle BAP is equal to angle ADP. We have done a simple construction also which is already shown in the figure. Here draw the diameter AOE and join AB. Now let us start with the triangle A, AB, angle EBA or angle ABE is equal to 90 degree since angle in a semicircle is of 90 degree. Therefore the sum of these two angles will be equal to 90 degree since the sum of three angles of a triangle is 180 degree and if one angle is 90 degree then the sum of remaining two angles is equal to 90 degree. So we have angle EB plus angle EAB is equal to 90 degrees. Let us be equation number one. Also we have A perpendicular on PQ so this implies angle EAQ is equal to 90 degrees or we have angle EAB plus angle BAQ is equal to 90 degrees. Let this be equation number two. Now from one and two we find that right hand side of both the equations are equal therefore left hand sides are also equal and angle AB is common on the left hand side so from one and two we have angle AEB is equal to angle BAQ and let this be equation number three. Now as we can see angle BEA and angle BCA are equal since that angle in the same segment of a circle that is angle AEB is equal to angle ACB since angles of a circle are equal and here the angles are in the same segment and let this be equation number four. Now from three and four we have left hand sides of both the equations are equal so this implies the right hand side is also equal so angle BAQ is equal to angle ACP. So this proves the first part and now we have to prove that angle BAP is equal to angle ADP. Now since PQ is the line therefore angle BAP plus angle BAQ is equal to 180 degrees linear pair also since opposite angles of a slightly correlated are supplementary therefore we have angle ACB plus angle ADB is equal to 180 degrees. Let this be equation number five and this be equation number six. Now since the right hand sides are equal therefore left hand sides are also equal so we have angle BAP plus angle BAQ equal to angle ACB plus angle ADP this is from equation number five and equation number six. Now angle BAQ is equal to angle ACB therefore this can further be written as angle BAP plus angle BAQ is equal to ACB can be written as BAQ plus angle ADB this further implies that angle BAP is equal to angle ADP so this proves the second part also. Now let us solve the given problem which is the second part of the question. Now here we are given RQ is equal to 40 degrees and angle PQR is equal to 80 degrees now by the above theorem if PQ is a chord then angle QPT is equal to angle QRP that is from the above theorem PQ is a tangent and sorry if PQ is a chord and ST is a tangent at the point P then angle QPT is equal to angle QRP right and angle QRP is given to us as 40 degrees so this implies angle QPT is equal to 40 degrees or angle X is equal to 40 degrees and also again from the above theorem we have for the chord RP angle RQP is equal to angle RTS and here again we are given that RQP is equal to 80 degrees so this implies angle RPS is equal to 80 degrees here since angle RQP is equal to 80 degrees that is again we get from the above theorem and which implies that Y is equal to 80 degrees hence our answer is X is equal to 40 degrees and Y is equal to 80 degrees so this completes the fashion buy and take care