Suppose you are an observer at O and there is an object Q, not in the same horizontal plane. Let OP be a horizontal line such that O, P and Q are in a vertical plane. Then if Q is above O, the angle QOP is the angle of elevation of Q observed from P, and if Q is below O, the angle QOP is the angle of depression.
Often when using angle of elevation and depression we ignore the height of the person, and measure the angle from some convenient 'ground level'.
Exercise 1: Opposite, Hypotenuse, Adjacent

Example 1: A Flagpole
From a point 10m from the base of a flag pole, its top has an angle of elevation of 50º. Find the height of the pole. [diagram] If the height is h, then ^{h}⁄_{10} = tan(50º). Thus h = 10 tan(50º) = 11.92m (to two decimal places). 
Example 2: A 15m High Flagpole
A flag pole is known to be 15m high. From what distance will its top have an angle of elevation of 50º? [diagram] If the distance is d, then ^{15}⁄_{d} = tan(50º). Thus d = ^{15}⁄_{tan(50º)} = 12.59m (to two decimal places). 
Example 3: A 20m Tower
From the foot of a tower 20m high, the top of a flagpole has an angle of elevation of 30º. From the top of the tower, it has an angle of depression of 25º. Find the height of the flagpole and its distance from the tower. Let the height of the flagpole be h and its distance be d. Then
The top of the flagpole is below the top of the tower, since it has an angle of depression as viewed from the top of the tower. It must be (20h) metres lower, so
Adding these two equations, we find
From this (how?) we find d = 19.16m and h = 11.06m (both to two decimal places). 
Example 4: Yet another Flagpole
From a certain spot, the top of a flagpole has an angle of elevation of 30º. Move 10m in a straight line towards the flagpole. Now the top has an angle of elevation of 50º. Find the height of the flagpole and its distance from the second point. [diagram] Let the height be h and its distance from the second point be x. Then
Subtracting the first expression from the second,

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