Two stations due south of a leaning tower which leans towards the north are at distances `aa n db` from its foot. If `alphabeta` be the elevations of the top of the tower from these stations, prove that its inclination `theta` to the horizontal is given by `cottheta=(bcotalpha-acotbeta)/(b-a)`

Two stations due south of a leaning tower which leans towards the north are at distances a and b from its foot If alpha and beta are the elevations of the top of the tower from these stations then prove that its inclination theta to the horizontal is given by cot theta=(b cot alpha-a cot beta)/(b-a)

Two stations due south of a leaning tower which leans towards the north are respectively at distance x and y from its foot (y gt x) . If alpha and beta be the elevations of the top of the tower from these stations and theta is inclination of the tower to the horizontal then cottheta equals

The respective ratio between the height of tower and the point at some distance from its foot is 5sqrt(3):5 .What will be the angle of elevation of the top of the tower ?

From a point on the ground which is at a distance of 50 m from the foot of the towe, the angle of elevation of the top of the tower is observed to be 30^(@) . Find the height of the tower.

From the foot of a building the elevation of the top of a tower is 45^(@) and from the foot of the twoer the elevation of the top of the building is 60^(@) . Which of the two is shorter ?

The angle of elevation of the top of a 15 meters high tower from a point 15 metres aways from its foot is:

The angle of elevation of the top of a tower from a point 40 m away from its foot is 60^(@) . Find the height of the tower.

C is the midpoint of the line joining two points A,B on the ground.A tower at C slightly leans towards B.If the angles of elevation of the top of the tower from A and B are 30^(@),60^(@) respectvely,the angle made by thetower with the horizontal is