**Introduction** How to apply trigonometric ratios, in the problems related with height and distance, you will know in this chapter. This chapter will enable you to measure, the width of river without crossing it, the height of a building without climbing it; so on many more measurements you will be able to do with the help of trigonometry. How? Let us know. Terms, which are essential to know before the actual start. Point of Observation: Whenever you see any object, your eye works as a point of observations. Though human beings can see an object with both eyes but we, for sake of convenience, consider it as we are watching the object with the single eye. The eye or the point from where you are watching the object is known as point of observation. In the pictorial representation of a height and distance problem this is represented by a single dot (point). There can be more than one point of observation you see an object from two or more than two points. Line of Sight: From the point of observation we see the object, it is supposed that we are not watching the object but only a particular point on it. It can be at the top of the object or at the bottom or anywhere else in the body of that object (according to the given problem). An imaginary line segment joining that point with the point of observation is known as line of sight or alternatively "line of vision". Angle of Elevation: The object which is in the consideration (in the sight) can be above or below the level of the eye (point of observation). At the level of point of observation we will draw a horizontal line. When the object is above the level of eye, then the angle which line of sight is making with this horizontal line is known as angle of elevation. Angle of Depression: When the object is below the level of point of observation, then the angle made by the line of sight with the horizontal line (drawn at the level of eye) is known as angle of depression. How to convert a height and distance problem in a line diagram? The best way to understand or to know the technique of conversion of height and distant problem is dealing with a few problems. Problem (I) A kite is flying at a height of 30m, from a point on the ground its angle of elevation is 60^{o}. Find the length of string.
The pictorial representation of this problem is the ABC. Problem (II) A man is 1.6m tall. He watches at the top of a tower which is 20m away from him and finds the angle of elevation 45^{o}. Find the height of the tower. The pictorial representation of this problem is the adjacent figure.
In the figure AB represents the height of the tower CD represent the height of man CE and DB represents the distance of man from the tower. Here AC represents the line of sight and ACE = 45^{o} represent the angle of elevation. Problem (III) A flag is mounted at the top of a building. From a point on the ground the angle of elevation of the top and bottom of the flag staff are 60^{o} and 45^{o}. If the building is 20m high then find the height of flag staff. The pictorial representation of this problem is the adjacent figure. In the figure AB represents the height of building. AD represents the flag staff. C represents the point of observation on the ground AC is the first line of sight and CD is another. ACB = 45^{o} and DCB = 60^{o} are the two angles of elevations. Problem (IV) Two cars are heading towards a tower through a straight road. From the top of the tower the angles of depression of these cars are 30^{o} and 30^{o}. If the height of the tower is 30m. Then find the distance between these cars.
The pictorial representation of this problem is the adjacent figure AB represents the height of tower.
Here A is point of observation C and D represents each car.
AX is the imaginary horizontal line.
AC and AD are the two lines of sight.
CAX and DAX are the two angels of depression DAX = 30^{o} and CAX = 60^{o}
AX and BD are the two parallel lines
CAX = ACB = 60^{o}
and XAD = ADB = 30^{o} [alternative angles] Problem (V) From the deck of a ship the angle of elevation of the top of a light house is 45^{o}. While the angle of depression of the bottom of the light house is 30^{o}. If the deck of the ship is 16m high then find the height of the light house.
The pictorial representation of this problem is the adjacent figure.
In the figure AB represents the light house.
CD represent the height of the deck of the ship C is the point of observation. CE the horizontal line which is situated at the level of point observation. BD is the ground level (level of sea). AC and CB are the two lines of sight ACE = 45^{o} is the angle of elevation ECB = 30^{o} is the angle of depression Since CE || BD ECB =CBD = 30^{o} |