ANGLE OF ELEVATION AND DEPRESSION

Angles of Elevation and Depression

Angle of Elevation:

Angle of elevation is the angle that an observer would raise his or her line of sight above a horizontal line to be able to view an object.

Angle of depression:

If an observer were up and is required to look down, the angle of depression would be the angle that the person would be required to lower his or her line of sight to be able to see an object.

The picture below shows an example of an angle of depression and an angle of elevation .

Assuming that you are looking at an object in the distance.
If the object is above you, then the angle of elevation is the angle your eyes look up.
If the object is below you, the angle of depression is the angle your eyes look down.
Angles of elevation and depression are measured from the horizontal.
It is a common mistake not to measure the angle of depression from the horizontal.

With the use of the angle of depression or elevation to an object, and knowing how far away the object is, we are enabled to find the height of the object through the use of trigonometry.

The advantage of doing this is that it is very difficult to measure the height of a mountain or the depth of a canyon directly; it is much easier to measure how far away it is (horizontal distance) and to measure the angle of elevation or depression.

Assuming that we want to calculate the height of this tree.
We mark point A and measure how far it is from the base of the tree.
Then we measure the angle of elevation from A to the top of the tree.

Now,

we have measured x and , therefore, we can calculate tan( ) and therefore we can find h, which is the height of the tree.

The angle of elevation of an object as seen by an observer is the angle between the horizontal and the line from the object to the observer’s eye (the line of sight).

If the object is below the level of the observer, then the angle between the horizontal and the observer’s line of sight is referred to as the angle of depression.

Example

From the top of a vertical cliff 40 m high, the angle of depression of an object that is level with the base of the cliff is 34º. How far is the object from the base of the cliff?

Solution:

Let x m be the distance of the object from the base of the cliff.

Therefore, the object is 59.30 m from the base of the cliff.

How to solve Questions on Angles of Elevation and Depression

To be able to solve problems that involves angles of elevation and depression, it is essential to

 Make use of basic right triangle trigonometry
 solve equations which involve one fractional term is as well very significant to know.
 Calculate an angle given a right triangle ratio of sides.
 the fact that corresponding angles formed by parallel lines have the same size.

A typical problem of angles of elevation and depression involves organizing information with regard to distances and angles within a right triangle. In a few cases, you will be asked to calculate the measurement of an angle; in others, the question might be to find an unknown distance.

Assuming a tree 50 feet in height casts a shadow of length 60 feet. What is the angle of elevation from the end of the shadow to the top of the tree with regard to the ground?

First we ought to make a diagram to organize our information. Look for these diagrams to involve a right triangle. In this case, the tree makes angle 90º with the ground. A diagram of this right triangle is shown below.

In the diagram, known distances are labeled. These are the 50 and 60 foot legs of the right triangle corresponding to the height of the tree and the length of the shadow.

The variable q is chosen to stand for the unknown measurement, the object of the question.

To relate the known distances and the variable, an equation is written. In this instance the equation involves the lengths of the sides which are opposite and adjacent to the angle q. With the use of the ratio of opposite to adjacent sides, we have .

We make use of inverse tangent of or which is the angle of elevation.

Trigonometry/Angles of Elevation and Depression

The diagram above is another representation of the angle of elevation and depression. Assuming you are an observer at O and there is an object Q, not in the same horizontal plane. Let OP be a horizontal line in a way 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.

Frequently, when making use of an angle of elevation and depression, we ignore the height of the person, and measure the angle from a few convenient ‘ground level’.

Exercise: Opposite, Hypotenuse, Adjacent

Look at the diagram above.

is a right angle. What would you give as the translation of the labels into English?

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?
If the distance is d, then 15⁄d = tan(50º). Therefore, 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
(i) h⁄d = tan(30º).
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 (20-h) metres lower, so
(ii) (20-h)⁄d = tan(25º).
Adding these two equations, we find
(iii) 20⁄d = tan(30º) + tan(25º).
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.

Solution

Let the height be h and its distance from the second point be x. Then
cot(50º) = x⁄h; cot(30º) = (x+10)⁄h
Subtracting the first expression from the second,
cot(30º)-cot(50º) = 10⁄h so
h = 10⁄(cot(30º)-cot(50º)) = 11.20 metres
x = hcot(50º) = 9.40 metres.

Example 5

Driving along a straight flat stretch of Abuja highway, you could see an especially tall saguaro (“suh-WARH-oh”) cactus right next to a mile marker. Watching your odometer, you pull over precisely two-tenths of a mile down the road. Retrieving your son’s theodolite from the trunk, you measure the angle of elevation from your position to the top of the saguaro as 2.4°. Accurate to the nearest whole number, how tall is the cactus?

Solution

Two-tenths of a mile is 0.2×5280 feet = 1056 feet, therefore this is my horizontal distance. You are required to find the height h of the cactus. You ought to draw a right triangle and label everything I know:

The scale is not significant; I’m not bothering to obtain the angle “right”. You would be making use of the drawing as a way to keep track of information; the particular size is irrelevant.

The essential thing is that you have “opposite” and “adjacent” and an angle measure. This means you can create and solve an equation:

h/1056 = tan(2.4°)
h = 1056×tan(2.4°) =
44.25951345…

To the nearest foot, the saguaro is 44 feet tall.

Question:

• You were flying a kite on a bluff, but you managed somehow to dump your kite into the lake below. You know that you’ve given out 325 feet of string. A surveyor tells you that the angle of declination from your position to the kite is 15°. How high is the bluff where you and the surveyor are standing?

Solution:

First, you would draw your triangle as shown below:

The horizontal line across the top is the line from which the angle of depression is measured. But by nature of parallel lines, the same angle is in the bottom triangle. You could easily observe the trig ratios more readily in the bottom triangle, and the height is a bit more clearly seen. Thus, you’ll make use of this part of the drawing.

You were given “opposite”, hypotenuse, and an angle,

so you’ll make use of the sine ratio to obtain the height.

h/325 = sin(15°)
h = 325×sin(15°) = 84.11618966…

The bluff is about 84 feet above the lake.

• A lighthouse stands on a hill 100 m above sea level.

If ∠ACD measures 60°and ∠BCD is 30°, find the height of the lighthouse.

Solution

You are supposed to work this exercise in steps. I can’t find the height of the tower, AB, until I have the length of the base CD. (Think of D as being moved to the right, to meet the continuation of AB, forming a right triangle.) For this computation, you’ll make use of the height of the hill.

100/|CD| = tan(30°)
100/tan(30°) = |CD| = 173.2050808…

To minimize round-off error, you’ll make use of all the digits from my calculator in my computations, and try to “carry” the computations in my calculator the whole way.

Now that you have the length of the base, you can calculate the total height, with the use of the angle that measures the elevation from sea level to the top of the tower.

h/173.2050808 = tan(60°)
h = 173.2050808×tan(60°) = 300

Brilliant! By keeping all the digits and carrying the computations in my calculator, you would get an exact answer. No rounding! But you do need to subtract, because “300” is the height from the water to the top of the tower. The first hundred meters of this total height is hill, thus:

The tower is 200 meters tall.