 Hello and welcome to the session. In this session we discuss the following question which says, from the top of a lighthouse the angles of depression of two ships on the opposite sides of it are observed to be theta and phi. If the height of the lighthouse, the edge meters and the line joining the ships passes through the foot of the lighthouse find the distance between the ships. So let's see how we do this. Consider this figure. Consider this figure where we have taken pq be the lighthouse. As in the question we are given that there are two ships on the opposite sides of the lighthouse. So we take let r and s be the positions of the ships and this r is the distance between the two ships which we have to find out. First of all, we would draw apb parallel to rs. So we have drawn apb parallel to rs. As in the question we are given that the angle of depression of two ships on the opposite sides of the lighthouse are theta and phi. So we take this angle apr as theta and angle bps as phi and we are also given that the height of the lighthouse is h meters. So we have pq is equal to h meters. Now since angle apr is theta so angle prq would be also theta. In the same way as angle bps is phi so angle psq would be phi. Since we have drawn apb parallel to rs so both these angles would be equal since they are the alternate interior angles. We also take let rq be equal to x meters then qs be equal to y meters. Now this lighthouse is the pendicular to the line joining the two ships. So we now consider the right triangle pqr. In this we have cot theta would be equal to base upon the pendicular that is x upon h which means that x is equal to h cot theta or you can say x is equal to h upon tan theta as we know that tan theta is equal to 1 upon cot theta. So cot theta is equal to 1 upon tan theta. Thus we have x is equal to h upon tan theta. Let this be equation one. Now next we would consider the right triangle pqs. In this we have cot phi is equal to base upon the pendicular that is y upon h which means y is equal to h cot phi or you can say that y is equal to h upon tan phi since we know that tan theta is equal to 1 upon cot theta. So cot theta is equal to 1 upon tan theta. Thus we get y is equal to h upon tan phi. Let this be equation two. Now since in the equation we have that we have to find the distance between the shapes and according to the figure we have r and s are the shapes. So the distance between the shapes is rs which is equal to x plus y and we have to find this. Now we have got the values for x and y. So now adding equations one and two we get plus y is equal to h upon tan theta plus h upon tan phi. For taking h common we have 1 upon tan theta plus 1 upon tan phi. Now taking LCM we have h into in the denominator we have tan theta into tan phi and in the numerator we have tan phi plus tan theta. So thus we say that x plus y is equal to h into tan theta plus tan phi this whole upon tan theta into tan phi. So we say that the distance between the two shapes is equal to x plus y which is equal to h into tan theta plus tan phi this whole upon tan theta into tan phi meters. So this is our final answer. With this we complete the session. Hope you have understood the solution of this question.