 Hi and welcome to the session. I am Shashi and I am going to help you with the following question. Question says the angle of elevation of the top of a building from the foot of the tower is 30 degrees and the angle of elevation of the top of the tower from the foot of the building is 60 degrees. If the tower is 50 meters high, find the height of the building. First of all let us understand what is the angle of elevation. The angle of elevation of an object viewed is the angle formed by the line of sight with the horizontal when it is above the horizontal level. Here the angle of elevation of P from Q is the angle formed by PQ with QR. QR is the horizontal line here. So we can say angle of elevation is equal to angle PQR. This is the key idea to solve the given question. Let us now start with the solution. First of all let us draw a simple diagram to represent the problem. Here let PQ be the tower and AB be the building. PQB is the horizontal distance between foot of the tower and foot of the building. Now we are given that the angle of elevation of top of a building from the foot of the tower is 30 degrees. Now this is the line of sight and this is the horizontal. So angle of elevation of top of a building is angle AQB and angle AQB is equal to 30 degrees. We are also given that the angle of elevation of the top of the tower from the foot of the building is 60 degrees. Here this is the line of sight and this is horizontal line. So angle of elevation of top of the tower is angle PBQ that is 60 degrees. Now we are also given that height of the tower that is PQ is equal to 50 meters. Now we have to find the height of the building. So we will find length of AB. First of all we will consider triangle AQB and find out horizontal distance QB. Now we are required to find height of the building. So we will find out length of AB. First of all we will consider triangle PQB and find out horizontal distance QB. Now let us start triangle PQB separately. Now in triangle PQB we know tan 60 degrees is equal to PQ upon QB. We know tan theta is equal to perpendicular upon wave. Here PQ is the perpendicular and QB is the wave. Now substituting corresponding values of tan 60 degrees and PQ in this expression we get root 3 is equal to 50 upon QB. Now multiplying both the sides by QB we get root 3 QB is equal to 50 dividing both the sides by root 3 we get QB is equal to 50 upon root 3 meters. Now we get QB is equal to 50 upon root 3 meters. Now let us consider triangle AQB. In triangle AQB we know QB is equal to 50 upon root 3 meters and angle AQB is equal to 30 degrees. Now let us draw this triangle separately. Now in triangle AQB tan 30 degrees is equal to AB upon BQ. We know tan theta is equal to perpendicular upon wave. Here AB is the perpendicular and BQ is the wave. Now substituting corresponding values of tan 30 degrees and BQ in this expression we get 1 upon root 3 is equal to AB upon 50 upon root 3. Now multiplying both the sides by 50 upon root 3 we get 1 upon root 3 multiplied by 50 upon root 3 is equal to AB. Now this implies 50 upon 3 is equal to AB. We know root 3 multiplied by root 3 is equal to 3 or we can simply write it as AB is equal to 50 upon 3 meters. Now this can be further written as 16 2 upon 3 meters. So height of the building is equal to 16 2 upon 3 meters. This is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.