 Hello and welcome to the session. In this session we discuss the following question which says the window of a house is h meters above the ground, the angles of elevation and the pressure of the top and the foot of another house from the window are alpha and beta respectively. Show that the height of the other houses h into 1 plus tan alpha into cot beta meters. Let's see how we do this. We have a house PQ with Q as its window. Now in the question we have that the window of this house is h meters above the ground. So PQ is equal to h meters. Now we have taken this RS as the other house. So first of all we will draw QT parallel to PR. So this QT is parallel to PR. In the question we have that the angle of elevation of the top of the other house from the window is alpha. So it means this angle is alpha as this is the top of the other house and this is the window of the house that we had taken. So this angle is alpha as we have angle SQT is equal to alpha and we are also given that the angle of depression of the foot of the other house from the window is beta. So it means this angle is beta that is angle TQR is equal to beta. Now as we have taken QP parallel to PR so this means angle TQR is equal to the angle QRP since they are the alternate interior angles and so this would be equal to beta that is angle QRP is also equal to beta. We take let RS be equal to h meters and here we have RT is also equal to this capital H and so TS would be equal to H minus capital H. Now we consider the right triangle QTS since this line is perpendicular to the house. So this is 90 degrees and this triangle QTS is right angle triangle. So in this right angle triangle we have cot alpha is equal to the base QT upon the perpendicular ST that is we have cot alpha is equal to QT upon ST which is OTS H minus H this gives us QT is equal to H minus H into cot alpha let this be equation 1. Now this triangle PQR is also right angle triangle which is right angle at P since this house would be perpendicular to the ground. So now we consider the right triangle QPR in this we take cot beta is equal to the base that is PR upon the perpendicular which is PQ this means we have cot beta is equal to PR upon PQ that is H or you can say that PR is equal to H into cot beta let this be equation 2 but we have that PR is equal to QT therefore from equations 1 and 2 we get H cot beta is equal to H minus H cot alpha that is we have H cot beta is equal to H cot alpha minus H cot alpha we get H cot alpha is equal to H into cot alpha plus cot beta this means we have H is equal to capital H into cot alpha plus cot beta this whole upon cot alpha which gives us H equal to capital H into 1 plus cot beta upon cot alpha or you can say we get H is equal to capital H into 1 plus tan alpha into cot beta since we know that tan theta is equal to 1 upon cot theta and we have taken the small H to be the height of the other house thus we say the height of the other house which is RS is equal to capital H into 1 plus tan alpha into cot beta meters this is what we were supposed to prove so hence proved so this completes the session hope you have understood the solution of this question