 Hello and welcome to the session. In this session we discuss the following question which says two stations do south of a leaning tower which leans towards the north at our distances c and d from its foot. If theta phi be the elevation of the top of the tower from these stations find its inclination to the horizontal. Let's proceed with the solution now. Consider this figure. Here we have taken let pq be the leaning tower and we have taken r and s to be the stations on the horizontal ground. These stations are towards the south of this leaning tower pq. It is also given that these stations are at distances of c and d from the foot of the leaning tower. So we have p r is equal to c and p s is equal to d. Theta and phi are the angles of elevations of the top of the star pq from the two stations. So we have angle qsp is equal to phi and angle qrp is equal to theta. First of all we would draw qt perpendicular to sx. So we have drawn this qt perpendicular to the horizontal ground sx. We take let the angle qpt be equal to alpha. Also we take let pt be equal to x and qt be equal to h. Now we consider the right triangle qpt in this triangle we have cot alpha is equal to the base that is pt upon the perpendicular which is qt. So cot alpha is equal to x upon h which means that x is equal to h cot alpha. We take this as equation 1. Next we consider the right triangle rqt. In this triangle we have cot theta is equal to the base rt upon the perpendicular qt which means cot theta is equal to c plus x upon h. Since we know that rt is equal to rp plus pt which is c plus x. So from here we get c plus x is equal to h cot theta that is x is equal to h cot theta minus c. Now from equation 1 we have x is equal to h cot alpha. So this means h cot alpha is equal to h cot theta minus c that is using equation 1. From here it follows h into cot alpha minus cot theta is equal to minus c. Or you can say h into cot theta minus cot alpha is equal to c which gives us h is equal to c upon cot theta minus cot alpha. Let this be equation 2. Next we consider the right triangle sqt. In this we have cot phi is equal to the base st upon the perpendicular qt which means cot phi is equal to d plus x since st is equal to sp plus pt that is d plus x upon qt which is h. This gives us d plus x is equal to h cot phi from where it follows that x is equal to h cot phi minus d. Again this means we have from equation 1 x equal to h cot alpha this means that we now get h cot alpha is equal to h cot phi minus d that is h into cot phi minus cot alpha is equal to d. This gives us h equal to d upon cot phi minus cot alpha. Let this be equation 3. So from equation 2 we have h equal to c upon cot theta minus cot alpha and from equation 3 we have h is equal to d upon cot phi minus cot alpha. Now from equations 2 and 3 we have c upon cot theta minus cot alpha is equal to d upon cot phi minus cot alpha. Now since in the equation we have to find the inclination of the leaning tower to the horizontal that is according to the figure we have the inclination of the leaning tower pq is alpha with the horizontal so we have to find out alpha. Now here cross multiplying we get c into cot phi minus cot alpha is equal to d into cot theta minus cot alpha which further gives us c into cot phi minus c into cot alpha is equal to d into cot theta minus d into cot alpha. This gives us c minus d into cot alpha is equal to c into cot phi minus d into cot theta. This further gives us cot alpha is equal to c into cot phi minus d into cot theta this whole upon c minus d. Now that we have to find the inclination of the leaning tower pq with the horizontal that is alpha. So we have now got cot alpha which means that alpha is equal to cot inverse of c into cot phi minus d into cot theta upon c minus d. So the inclination of the leaning tower with the horizontal is given by alpha equal to cot inverse c cot phi minus d cot theta upon c minus d. This is our final answer. This completes the session hope you have understood the solution of this question.