 Hello and welcome to the session. The given question says if the angle of elevation of a cloud from a point h meters above a lake is alpha and the angle of depression of its reflection in the lake is beta, prove that the distance of the cloud from the point of observation is 2h6 alpha divided by tan beta minus tan alpha. So in this figure suppose this line x y denotes the base or the level of the ground, a person is standing h meters above the lake and is making an angle of alpha watching this cloud. Now suppose this distance bbb equal to capital H meters and the height at which the observer is standing is h meters above the level of the lake. Now here the distance cp dash is equal to the distance pc since the image of an object below the c level is equal to the image of the object above the c level. So if this is small h meters this is capital H meters so the distance cp dash will be capital H plus a small h meters. Let's now start with the solution. Now in triangle a pb we have bp divided by ap is equal to sin alpha which implies that h divided by ap is equal to sin alpha or we have ap is equal to h divided by sin alpha. Let us denote this by equation number one. We have to prove that the distance of the cloud from the point of observation is 2h6 alpha divided by tan beta minus tan alpha. That is we have to prove that the distance ap is equal to 2h6 alpha divided by tan beta minus tan alpha. Now here we had assumed that pb is equal to capital H meters bc is equal to small h meters therefore cp dash is equal to the distance pb plus bc that is capital H plus small h meters since the distance of an object above the c level is equal to the distance of an object below the c level. Now let us consider triangle app dash. Now in triangle a bp dash we have bp dash divided by ap is equal to tan beta. Now bp dash is the distance bc plus cp dash now bc is h and cp dash is h plus small h. So we have h plus two times of small h divided by ap is equal to tan beta and again if we see triangle apb then here pb divided by ap is equal to tan alpha so this implies that ap is equal to pb divided by tan alpha. So here we have capital S plus two times of small h is equal to ap into tan beta and ap is pb divided by tan alpha into tan beta. Now let us substitute the value of pb which is capital H. So we have h plus two h is equal to capital H into tan beta divided by tan alpha. So this implies that two h is equal to capital H into tan beta divided by tan alpha minus one or we have capital H equal to two h into tan alpha whole divided by tan beta minus tan alpha and let this be equation number two. Now we have to show that ap is equal to two h say alpha divided by tan beta minus tan alpha. So let us substitute the value of capital H in equation number one. So substituting the value of capital H in equation number one we have ap is equal to h divided by sin alpha and capital H is two h tan alpha divided by tan beta minus tan alpha and this divided by sin alpha and this is equal to two h tan alpha is sin alpha divided by cos alpha divided by tan beta minus tan alpha this divided by sin alpha and this is equal to two h divided by cos alpha since sin alpha cancels off with sin alpha and in the denominator we have tan beta minus tan alpha and this is equal to two h sin alpha whole divided by tan beta minus tan alpha. Therefore we have ap is equal to two h sin alpha whole divided by tan beta minus tan alpha and thus we have proved that the distance of the cloud point of observation alpha divided by tan beta minus tan alpha. So this completes the session. Bye and take care.