Angles of Elevation and Depression are used in measuring heights and distances in trigonometric applications using right triangles. These angles are made when we look up or down to view objects. Devices are available to measure angles of elevation and depression. These measured angles can be used in measuring heights and distance which are either tedious or impractical to measure, by modelling the situation into right triangles.

In the above picture a man standing on a cliff is simultaneously viewing an airplane flying in the sky and a boat sailing in the lake beneath. The angles of elevation and depression are marked as xº and yº respectively.

## Angle of Elevation

Angle of elevation is the angle formed by the line of sight of an observer with the horizontal while he is viewing an object upward.

The angle of elevation is a degree measure indicating the gradient of the line of sight. As this angle can be measured using devices, this measure can be used to determine large heights and long distances, which are impractical to physically measure.

Elevation of Sun is a degree measure similar to angle of elevation, which represents the the distance of the center of Sun's disk from the Horizon. Elevation of Sun is also used frequently in Right triangle model applications.

In the above diagram Sun's elevation is marked as the angle between the line joining the Sun and the tip of the shadow and the horizontal.

## Angle of Depression

When the object of sight falls below the horizontal at the eye level, an angle of depression is formed.
Definition: Angle of depression is the angle formed by the line of sight of an observer with the horizontal while he is viewing an object downward.

Generally an angle formed with the ground level horizontal, which is the alternate interior angle of the angle of depression is used in right triangle models to solve application problems. In the above illustration, you may note the alternate interior angle at the ground level.

Even though we normally consider alternate interior angle for solving the problems, the angle of depression has to be shown different from this in the sketch. The wrong and right sketches are given below.

## Angles of Elevation and Depression Formula

Angles of Elevation or Depression can be got using inverse trigonometric functions if the related lengths are known. In the right triangle model, if θ is a measure of angle of elevation or depression then,

θ = sin-1 ($\frac{Opposite\ Leg}{Hypotenuse}$)

or

θ = cos-1 ($\frac{Adjacent\ Leg}{Hypotenuse}$)

or

θ = tan-1 ($\frac{Opposite\ Leg}{Adjacent Leg}$)

Use the appropriate formula depending on the known measures in the triangle.

### Solved Example

Question: The height of a building is 250 ft. What is the angle of elevation from a point on the level ground 200 ft away from the base of the building?

Solution:

The lengths of opposite and the adjacent legs are known for angle θ which is the angle of elevation.

θ =  tan-1 ($\frac{Opposite\ Leg}{Adjacent\ Leg}$) = tan-1  $\frac{250}{200}$

≈ 51.3º.           Using inverse functions in the calculator.

## Angle of Elevation Vs Angle of Depression

Let us compare the characteristics of angles of elevation and depression.

 Angle of Elevation Angle of Depression Angle of elevation is measured when the object is viewed upward. Angle of depression is measured when theobject is viewed downward. Angle of elevation is always an angle in the right triangle considered for solving theproblem Generally, an angle in the right triangle whichis the alternate interior angle of the angle ofdepression is considered for solving the problem. When two positions are each viewed from the other position then, the angle of elevation from one point is equal to the angle of depression from the other point, as two angles are alternate interior angles.

## Angle of Elevation and Depression Problems

Let us solve few word problems involving angles of elevation and depression.

### Solved Examples

Question 1: A surveyor wanted to measure the height of a mountain. He traveled in the straight road leading to the mountain and measured the angle of elevation of the peak from a point A which is on level ground with the mountain base as 32º. He drove 1 Km further down the road and again measured the angle of elevation of the peak as 40º. Find the height of the mountain nearest to a meter.

Solution:

As we see in the pictures, two variables h and x are introduced, the height of the mountain and the distance BC. Using trig formulas we can eliminate x and solve for h.
In right  triangle BCD,

tan 40 = $\frac{h}{x}$   ⇒       x = $\frac{h}{tan 40}$

in right triangle ACD,

tan 32 = $\frac{h}{x+1}$

= $\frac{h}{\frac{h}{tan40}+1}$                 Substitution

= $\frac{h\tan40}{h+tan40}$                      Complex fraction simplified.

(h + tan 40) tan 32  = htan 40                              Cross multiplication
htan 32 + tan 40.tan 32 = htan40
htan 40 - htan 32 = tan 40.tan 32
h(tan 40 - tan 32) = tan 40.tan 32

h = $\frac{tan40tan32}{tan40-tan32}$

≈ 2.4475 Km
Hence the estimated height of the mountain = 2447.5 meters.

Question 2: Roger views from his shop window which is 20 ft above ground level, the top and the base of a building on the opposite side of the Road. If the angles of elevation and depression of the top and base are 64º and 28º find the height of the building he is viewing rounded to the tenth of a foot.

The situation can be sketched as follows:

Solution:

AB represents the height at which Roger is viewing and = 20 ft. CD represents the building he is viewing and we need to find its measure which is equal to (20 + y) ft.
In right triangle AEC,

tan 28º = $\frac{CE}{AE}$ = $\frac{20}{x}$

x = $\frac{20}{tan28}$ ≈ 37.61 ft

In right triangle AED,

tan 64º = $\frac{y}{x}$ = $\frac{y}{37.61}$

y = 37.61 x tan 64  ≈ 77.12 ft.

Height of the building = 77.1 + 20 = 97.1 ft   (rounded to the tenth of a foot).

Question 3: The shadow of a tower, when the angle of elevation of the Sun is 30º is found to be 40 meters longer than when it is 45º. Find the height of the Tower.

Solution:

The above sketch depicts the situation.
The height of the tower is represented by AB which is taken as y ft.
ABC is an isosceles right triangle ⇒  y = x.
In right triangle ABD,

tan 30º = $\frac{y}{x+40}$ = $\frac{x}{x+40}$                          Definition of tangent of an angle.

x = (x + 40)tan 30                                                                    Cross multiplication.
x = x tan 30 + 40 tan 30
x- x tan 30 = 40 tan 30

x = $\frac{40tan30}{1-tan30}$ ≈ 54.6 ft.