ABC is a vertical tower such that AB=BC=x from the foot A of the tower a horizontal line AQP is drawn.If the angle of elevation of the point B from P is `alpha` and that of C from Q is `2 alpha` then `(AP)(PQ)` is equal to

If the distance of a point from the tower is equal to the height of the tower , then find the angle of elevation of the top of the tower from that point .

PQ is a vertical tower having P as the foot. A,B,C are three points in the horizontal plane through P. The angles of elevation of Q from A,B,C are equal and each is equal to theta . The sides of the triangle ABC are a,b,c, and area of the triangle ABC is . Then prove that the height of the tower is (abc) tantheta/(4)dot

The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60^(@) and the angle of elevation of the top of the second tower from the foot of the first tower is 30^(@). The distance between the two towers is m times the height of the shorter tower. What is m equal to ?

From a point on the ground,20m away from the foot of a vertical tower,the angle of elevation of the top of the tower is 60o, what is the length of the tower?

The angles of depression of two points A and B on a horizontal plane such that AB= 200 from the top P of a tower PQ of height 100 are 45 - theta and 45 + theta . If the line AB Passes through Q the foot of the tower, then angle theta is equal to

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.