 Hello and welcome to the session. In this session, we will discuss a question that says that in figure, pq and pr are tangents from an external point p to the circle with center o. If every rpq is equal to 13 degrees, won't that open is equal to pq? By the first adding resolution of this question, we should know some results. First is, center lies on the bus sector of the angle between the two tangents. That means that if r is the center of the circle and a p and a c are the tangents, then every lie on the bus sector of the angle be a c. That is o a by 6 angle be a c, which implies angle b a o is equal to angle o a c. Secondly, tangent is perpendicular to the point of contact. Now these results will work out as a t i here for solving out this question. And now, we will start with the solution. Now given, pq and pr are the tangents of the circle with center o. Now here, we have particular construction that is drawing oq. Now oq is the radius as pr is the center of the circle. Now using the second result, which is given in the t i here, angle oq b is equal to 90 degrees because tangent is perpendicular to the radius through the point of contact as oq is the radius and pq is the tangent and q is the point of contact. Therefore, using this result, angle oq b is equal to 90 degrees. Now using the first result, which is given in the q i here, Now we have o p by sets angle r pq because on the bisector, so this implies equal to angle o pq. Now you can be written as angle r p o plus angle o pq. Which further implies angle r pq is equal to, now here, angle r p o is equal to angle o pq. So replacing this value, it will be angle o pq plus angle o pq. 2 is equal to 2 into angle o pq. Now it is given that angle r pq is equal to 120 degrees. So putting this value here, this implies 120 degrees is equal to 2 into angle o pq. This further implies angle o pq is equal to 120 degrees by 2 which is equal to 60 degrees. Now this angle is 60 degrees. Now in triangle o pq, 60 degrees is equal to base over hyper-nose and here base is pq over hyper-nose is o p. Now this implies, now past 60 degrees is 1 by 2 is equal to pq by o p. Which further implies on past multiplying o p is equal to 2 into pq. Now to prove, o p is equal to 2 into pq, the solution of the given question. And that's all for this session. Hope you all have enjoyed the session.