Angles of Elevation & Depression

Angle of Elevation:

This is the angle between normal eye level (horizontal line) and the line through which an object is above the observer’s eye.

Angle of Depression:

The angle of depression is the angle formed when an observer looks down at an object. It’s the angle between the horizontal line from the observer’s eye level and the line of sight to the object below

Relationship between Angle of Elevation and Depression:

From the diagram, the horizontals are parallel, hence, 

e = d (Alternate angles) 

⇒ Angle of elevation = Angle of depression 

The angles of elevation and depression are mostly applicable in solving problems involving heights and distances.

Example 6.1.1:

Two observers, Abu and Badu, 46 m apart, observe a bird on a vertical pole from the same side of the bird. The angles of elevation of the bird from Abu’s and Badu’s eyes are 40° and 48 °, respectively. If at the foot of the pole, Abu and Badu are on the same horizontal;

(a) Illustrate the information in a diagram
(b) Calculate, correct to one decimal place, the height of the pole

Solution

(a)

(b)

consider ΔCBD

tan θ = \( \frac{opposite}{adjacent}\\ = \frac{CD}{BD}\)

∴ tan 48º = \(\frac{h}{x} \)

h = x tan 48º …….(1)

consider ΔCAD

tan θ = \( \frac{opposite}{adjacent}\\ = \frac{CD}{AD}\)

∴ tan 40º = \(\frac{h}{46 \: + \: x} \)

h = (46 + x) (tan 40º) …….(2)

equating 1 and 2

x × 1.11 = (46 + x) (0.839)

1.11x = 38.594 + 0.839x

1.11x – 0.839x = 38.594

0.271x = 38.594

x = \( \frac{38.594}{0.271}\)

x = 142.413

but h = x tan 48º …….(1)

∴ h = 142.413 × tan 48º

h = 142.413 × 1.11

h = 158.1 m (1 d.p.)

Example 6.1.2:

A boy stands at a point, M, on the same horizontal level as the foot, T, of a vertical building. He observes an object on the top, P of the building at an angle of elevation of 66°. He moves directly backwards to a new point C and observes the same object at an angle of 53°.

If |MT| = 50 m:

(a) Illustrate the information in a diagram;
(b) Calculate and correct to one decimal place: 
(i) the height of the building;
(ii) \( \scriptsize \overline {|MC|} \)

Solution:

Solution

(b) Calculate and correct to one decimal place:

(i) the height of the building;

Solution

tan θ = \( \frac{opp}{adj}\)

tan 66° = \( \frac{h}{50}\)

h =  tan 66° × 50

h = 112.3018

h = 112.3 m

(ii) \( \scriptsize \overline {|MC|} \)

Solution

Considering ΔPCT

⇒ tan 53° = \( \frac{\overline{PT}}{\overline{CT}}  \)

⇒ \(\scriptsize \overline{CT} = \normalsize \frac{\overline{PT}}{tan \:53^o}\)

⇒ \(\scriptsize \overline{CT} = \normalsize  \frac{112.3}{1.327}\)

⇒ \(\scriptsize \overline{CT} = 84.627\: m\)

⇒ \(\scriptsize \overline{CT} = \overline{CM} \: + \: \overline{MT}\)

⇒ \(\scriptsize 84.627\: m = \overline{CM} \: + \: 50\: m\)

⇒ \(\scriptsize \overline{CM} = 84.627\: m  \: – \: 50\: m\)

⇒ \(\scriptsize \overline{CM} = 34.627 \: m \)