Angle of Depression

Let O be the eye of an observer and A be an object below the level of the eye. The ray OA is called the line of sight. Let OB be the horizontal line through O. Then the angle BOA is called the angle of depression of the object A as seen from O.

Angle of Depression

It may so happen that a man climbs up the pole, keeps his eyes at a point O and see the object placed at the point A is the angle of depression of the point A with respect to the point O.

 How can we get the angle of depression?

Angle of Depression Image

We shall have to imagine a straight line OB parallel to the straight line CA. The measure of the angle of depression will be ∠BOA.


It is clear from the figure below that the angle of elevation of A as seen from B = the angle of depression of B as seen from A.

Angle of Elevation and Angle of Depression

Therefore, ∠θ = ∠β.

Note: 1. Here, BC ∥ DA and AB is the transversal. So the angle of elevation ∠ABC = the angle of depression ∠BAD. But even then they are to be indicated to solve problems.

2. The observer is taken as a point unless the height of the observer is given.

3. √3 = 1.732 (Approximately).


10th Grade Heights and Distances

Solved Examples on Angle of Depression:

1. From the top of a tower, a man finds that the angle of depression of a car on the ground is 30°. If the car is at a distance 40 metres from the tower, find the height of the tower.

Solution:

Let PQ be the tower and the car is at R. 

The angle of depression = ∠SPR = 30° and QR = 40 m.

From geometry, ∠PRQ = ∠SPR = 30°.

Problems on Angle of Depression

In the right-angled ∆PQR, 

                           tan  30° = \(\frac{PQ}{QR}\)

                       ⟹ \(\frac{1}{√3}\) = \(\frac{PQ}{40 m}\)

                       ⟹ √3PQ = 40m

                       ⟹ PQ = \(\frac{40}{√3}\) m

                       ⟹ PQ = \(\frac{40√3}{3}\) m

                       ⟹ PQ = \(\frac{40 × 1.732}{3}\) m

                       ⟹ PQ = 23 m (Approx.).

Therefore, the height of the tower is 23 m (Approx.).


Angle of Depression Example 

2. From the top of a cliff 200 m height, the angles of depression of two places A and B on the ground and on the opposite sides of the cliff are 60° and 30°. Find the distance between M and N.

Solution:

Let TO be the cliff, and given that TO = 200 m.

M and N are the two points.

The angle of depression ∠X'TM = 60° and ∠XTN = 30°.

By geometry, ∠TMO = 60° and ∠TNO = 30°.

Problems on Angle of Depression

In the right-angled ∆TOM, 

                                 tan 60° = \(\frac{TO}{MO}\)

                            ⟹       √3  = \(\frac{200 m}{MO}\)

                            ⟹       MO  = \(\frac{200 m}{√3}\)


In the right-angled ∆TON, 

                                 tan 30° = \(\frac{TO}{NO}\)

                            ⟹       \(\frac{40}{√3}\) = \(\frac{200 m}{NO}\)

                            ⟹       NO  = 200√3 m.

Therefore, the required distance MN = MO + NO 

                                                     = \(\frac{200 m}{√3}\) + 200√3 m.

                                                     = \(\frac{200 + 600}{√3}\) m

                                                     = \(\frac{800}{√3}\) m

                                                     = \(\frac{800√3}{3}\) m

                                                     = \(\frac{800 × 1.732}{3}\) m

                                                     = 461.89 m (Approx.)


Word problems on Angle of Depression:

3. A building stands on the bank of a river. A man observes from a corner of the roof of the building, the foot of a electric post just on the opposite bank. If the angle of depression of the foot of the light post at your eye is 30° and the height of the building is 12 meters, what is the width of the river?

Solution:

Let P is the roof of the building, Q is the foot of the building vertically below the corner point and R is the foot of the light post just on the opposite of the bank of the river. A right-angled triangle PQR is formed by joining these points.

Angle of Depression Heights and Distances

Let PS be the horizontal line through P.

∠SPR, the angle of depression = ∠PRQ = 30°, and with respect to this angle perpendicular PQ = 12 metres and base QR = width of the river = h metres.

From right-angled triangle PQR,

\(\frac{PQ}{QR}\) = tan 30°

\(\frac{12}{h}\) = \(\frac{1}{√3}\)

⟹ h = 12 × √3

⟹ h = 12 × 1.732

⟹ h = 20.784 (Approximately)

Therefore, the width of the river is 20.784 meters (Approximately).


Angle of Depression Problem:

4. From the top of a building, the angle of depression of the top and the foof of a lamp post are 30° and 60° respectively. What is the height of the lamp post?

Solution:

According to the problem, the height of the building PQ = 12 m.

Let height of the lamp post RS.

Angle of depression of the top of a lamp post is 30°

Therefore, ∠TPR = 30°.

again, Angle of depression of the foot of a lamp post is 60°

Therefore, ∠TPS = 60°.

PQ = TS = 12 m.

Let the height of the lamp post RS = h m.

10th Grade Heights and Distances

Therefore, 

TR = (12 - h) m.

Also, let PT = x m

Now tan ∠TPR = \(\frac{TR}{PT}\) = tan 30°

Therefore, \(\frac{12 - h}{x}\) = \(\frac{1}{√3}\) ........... (i)

Again, tan ∠TPS = \(\frac{TS}{PT}\) = tan 60°

Therefore, \(\frac{12}{x}\) = √3 ........... (ii)

Dividing (i) by (ii), we get

\(\frac{12 - h}{12}\) = \(\frac{1}{3}\)

⟹ 36 - 3h = 12

⟹ 3h = 36- 12

⟹ 3h = 24

⟹ h = \(\frac{24}{3}\)

⟹ h = 8

Therefore, height of the lamp post is 8 metres.




10th Grade Math

From Angle of Depression to HOME


New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.



Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



Share this page: What’s this?

Recent Articles

  1. Types of Fractions |Proper Fraction |Improper Fraction |Mixed Fraction

    Mar 02, 24 05:31 PM

    Fractions
    The three types of fractions are : Proper fraction, Improper fraction, Mixed fraction, Proper fraction: Fractions whose numerators are less than the denominators are called proper fractions. (Numerato…

    Read More

  2. Subtraction of Fractions having the Same Denominator | Like Fractions

    Mar 02, 24 04:36 PM

    Subtraction of Fractions having the Same Denominator
    To find the difference between like fractions we subtract the smaller numerator from the greater numerator. In subtraction of fractions having the same denominator, we just need to subtract the numera…

    Read More

  3. Addition of Like Fractions | Examples | Worksheet | Answer | Fractions

    Mar 02, 24 03:32 PM

    Adding Like Fractions
    To add two or more like fractions we simplify add their numerators. The denominator remains same. Thus, to add the fractions with the same denominator, we simply add their numerators and write the com…

    Read More

  4. Comparison of Unlike Fractions | Compare Unlike Fractions | Examples

    Mar 01, 24 01:42 PM

    Comparison of Unlike Fractions
    In comparison of unlike fractions, we change the unlike fractions to like fractions and then compare. To compare two fractions with different numerators and different denominators, we multiply by a nu…

    Read More

  5. Equivalent Fractions | Fractions |Reduced to the Lowest Term |Examples

    Feb 29, 24 05:12 PM

    Equivalent Fractions
    The fractions having the same value are called equivalent fractions. Their numerator and denominator can be different but, they represent the same part of a whole. We can see the shade portion with re…

    Read More