 Hello and welcome to the session. In this session, we discuss the following question which says the angle of elevation of a cloud from a point h meters above a lake is theta and the angle of depression of its reflection in the lake is phi proves that the height of the cloud is h into tan phi plus tan theta upon tan phi minus tan theta meters. Let's see its solution now. We take, let pq be the surface of the lake and we have taken a point r vertically above the point p such that we have rp is equal to h meters as in the question is given that there is a point h meters above the lake. So we have taken that point as point r and we have rp is equal to h meters then we have taken let a be the position of the cloud and let b be the reflection of the cloud in the lake we assume let the height of the cloud be h meters then we have aq is equal to qb is equal to h meters this is h and this is h we have taken this r as perpendicular to ab now this rp is h and this aq is h so this as would be h minus h from the question we have the angle of elevation of the cloud from the point which is h meters above the lake is theta so it means that this angle is theta that is we have angle ars is equal to theta also we have that the angle of depression of the reflection of the cloud is phi so it means this angle is phi thus angle brs is equal to phi consider the right triangle ars in this we have cot theta is equal to the base which is ars upon the perpendicular that is as which means we have cot theta is equal to ars upon as which is h minus h this gives us ars is equal to h minus h into cot theta we take this as equation one now let's take the right triangle brs in this right triangle we have cot phi is equal to the base that is ars upon the perpendicular which is bs it means we have cot phi is equal to ars upon bs now bs is equal to bq plus qs now bq is h plus qs which is equal to rp that is h so h plus this capital H is bs so cot phi is equal to ars upon h plus h from here we get ars is equal to h plus h into cot phi let this be equation 2 so from equations 1 and 2 we have h minus h cot theta is equal to h plus h cot phi this means we have cot theta upon cot phi is equal to h plus h upon h minus h or you can say we have tan phi upon tan theta is equal to h plus h upon h minus h since we know that tan theta is equal to 1 upon cot theta now by component low dv dendo we get tan phi plus tan theta upon tan phi minus tan theta is equal to h plus h plus h minus h upon h plus h minus h minus h so we have tan phi plus tan theta upon tan phi minus tan theta is equal to 2h upon 2h so from here we get h upon this capital H is equal to tan phi plus tan theta upon tan phi minus tan theta this means we have h is equal to capital H into tan phi plus tan theta upon tan phi minus tan theta and we have taken this edge to be the height of the cloud so we say height of the cloud is equal to h into tan phi plus tan theta upon tan phi minus tan theta meters this is what we were supposed to prove so hence proved this completes the session hope you have understood the distribution of this question