 Hello and welcome to the session. In this session we discussed the following question that says two tangents PQ and PR are drawn to a circle with center O such that angle QPR is equal to 60 degrees proves that 2 times OQ is equal to root 3 times OP. Before we move on to the solution, let's recall the RHS congruence rule. According to this we have that if in two triangles, hypotenuse and one side of a triangle are equal to the hypotenuse and one side of other triangle, then the two triangles are congruent. This is the key idea that we use in this question. Now let's see the solution. Consider this figure. In this, we are given that PQ and PR are the tangents drawn to the circle and also we are given that angle QPR is equal to 60 degrees and we need to prove that 2 times OQ is equal to root 3 times OP. Now since we know that the tangent at any point of a circle is perpendicular to the radius through the point of contact, therefore this would mean that angle OQP is equal to angle ORP and each is equal to 90 degrees. So this means triangles OQP and ORP are right triangles. Now consider the right triangles OPQ and OPR. In these two triangles we have OQ is equal to OR as they are the radii of the same circle then OP is equal to OP. It is the common side. So therefore triangle OPQ is congruent to the triangle OPR by the RHS congruence rule stated in the key idea. Now since both these triangles are congruent, so this would mean that angle OPQ is equal to angle OPR as they are the corresponding parts of congruent triangle. So they are equal. That is these two angles would be equal. It's given to us that angle QPR is equal to 60 degrees and we know that angle QPR is equal to angle QPO plus angle OPR and both these angles are equal. So this would mean angle QPO is equal to angle OPR is equal to 60 degrees upon 2 that is equal to 30 degrees. So this is 30 degrees. This is 30 degrees. Now in triangle OQP angle OQP plus angle QPO plus angle QOP is equal to 180 degrees by the angles and property of triangle. So this means 90 degrees that is angle OQP is 90 degrees plus angle QPO which is 30 degrees plus angle QOP is equal to 180 degrees. This means angle QOP is equal to 180 degrees minus 120 degrees equal to 60 degrees. So now you can say that sine of 60 degrees is equal to the base that is OQ upon the hypotenuse that is OP. Now sine 60 degrees is root 3 upon 2 is equal to OQ upon OP which means that 2 times OQ is equal to root 3 times OP. So we were supposed to prove this hence proved. This can be easy session. Hope you have understood the solution of this question.