Height is the measurement of an item in the vertical direction, whereas distance is the measurement of an object in the horizontal direction. Heights and Distances are the real-life applications of trigonometry which is useful to astronomers, navigators, architects, surveyors, etc. in solving problems related to heights and distances.
In height and distance, we use trigonometric concepts to find the height and distance of various objects.
Table of Contents
Height and Distance in Trigonometry
Various terminologies that help understand Height and Distance are,
- Line of Sight: It is the line drawn from the eye of an observer to the point in the object viewed by the observer.
- Angle of Elevation: The angle between the horizontal and the line of sight joining an observation point to an elevated object is called the angle of elevation.
- Angle of Depression: The angle between the horizontal and the line of sight joining an observation point to an object below the horizontal level is called the angle of depression.
How to Find Heights and Distances?
Trigonometric ratios are used to measure the heights and distances of different objects. For example, in the above figure, a person is looking at the top of the tree the angle from the eye level to the top of the tree is called the angle of elevation and similarly, the angle from the top of the tree to the eyes of the person is called the angle of depression.
If the height of the person and its distance from the tree is known we can easily calculate the height of the tree using various trigonometric ratios.
To measure heights and distances of different objects, we use trigonometric ratios. For example, in fig.1, a guy is looking at the top of the lamppost. AB is horizontal level. This level is the line parallel to ground passing through the observer’s eyes. AC is known as the line of sight. ∠A is called the angle of elevation. Similarly, in fig. 2, PQ is the line of sight, PR is the horizontal level and ∠P is called the angle of depression.
Angle of Elevation in Height And Distance
Angle of Depression in Height And Distance
The angles of elevation and depression are usually measured by a device called Inclinometer or Clinometer.
Inclinometer for Measuring Angles of Elevation and Depression
Trigonometric Ratios Table
The value of trigonometric ratios for different angles is very useful for solving Height and Distance problems. Thus it is advised to learn the values of trigonometric ratios for different angles. The value of various trigonometric ratios can be learned using the trigonometric table provided below,
| Angles (In Degrees) | 0 | 30 | 45 | 60 | 90 | 180 | 270 |
|---|---|---|---|---|---|---|---|
| Angles (In Radians) | 0 | π/6 | π/4 | π/3 | π/2 | π | 3π/2 |
| sin | 0 | 1/2 | 1/√2 | √3/2 | 1 | 0 | -1 |
| cos | 1 | √3/2 | 1/√2 | 1/2 | 0 | -1 | 0 |
| tan | 0 | 1/√3 | 1 | √3 | Not Defined | 0 | Not Defined |
| cot | Not Defined | √3 | 1 | 1/√3 | 0 | Not Defined | 0 |
| cosec | Not Defined | 2 | √2 | 2/√3 | 1 | Not Defined | -1 |
| sec | 1 | 2/√3 | √2 | 2 | Not Defined | -1 | Not Defined |
Heights and Distances Formulas
In this section, you can see three different cases when solving height and distance questions.
Case 1: In this case, we can observe the following:
- Height of a tower, hill or building
- Distance of an object from the foot of the tower, hill or building and sometimes shadow of them
- Angle of elevation or the angle of depression
Any two of the above three parameters will be provided in the question. This type of problems can be solved using the formulas given below.
In right triangle ABC,
sin θ = Opposite/Hypotenuse = AB/AC
cos θ = Adjacent/Hypotenuse = BC/AC
tan θ = Opposite/Adjacent = AB/BC
Case 2: In this case, we can deal with different illustrations. One of the commonly solved problems is about the movement of an observer. If the observer moves towards the objects like a tower, building, hill, etc., then the angle of elevation increases. The angle of elevation decreases when the observer moves away from the object. Here, the distance moved by the observer can be found using the formula given below:
In right triangle given below, d is the distance between C and D.
d = h(cot x – cot y)
Case 3: There is another case where two different situations happening at the same. In this case, we get similar triangles with the same angle of elevation or angle of depression. These type of problems can be solved with the help of formulas related to similar triangles.
In right triangle ABC, DE || AB,
Here, triangles ABC and EDC are similar. Using Thales or BPT theorem we can write the ratio of sides as:
AB/ED = BC/DC
The above three formulas are used to solve majority of the heights and distances problems. Apart from these cases, there some other types of constructions could be drawn based on the given scenario. However, the combination of these formulas are used to solve any given problem related to the heights and distances in trigonometry.
Heights and Distances (Case Study)
Question: An aeroplane is flying h meters above the ground. At a particular instant, the angle of elevation of the plane from the eyes of a boy sitting on the ground is 60°. After some time, the angle of elevation changed to 30°. Find the distance covered by the plane during that time assuming it travelled in a straight line.
Solution:
The problem explained in the question can be drawn as shown in the figure.
Height And Distance Example Question
In ∆ OAB,
tan 60° = AB/OA
√3 = h/x
x = h/√3
In ∆ OCD,
tan 30° = CD/OD
1/√3 = h/(x+y)
x + y = √3h
Distance travelled by plane = AD = y
(x + y) − x = √3h − h/√3
y = (2/√3)h
So, if the aeroplane is flying h meters above the ground, it would travel for (2/√3) h meters as the angle of elevation changes from 60° to 30 °.
Solved Examples on Height and Distance
Example 1: If a pole 6 m high casts a shadow 2√3 m long on the ground, find the Sun’s elevation.
Solution:
Let AB be the pole which is of height 6 m and BC be the shadow of the building 2√3.
Now, in ∆ ABC,
tan θ = AB / BC
⇒ tan θ = 6 / 2√3Now, simplifying using rationalization
tan θ = (3 / √3) × (√3 / √3)
⇒ tan θ = 1 / √3
⇒ θ = tan-1(1 / √3)Hence, θ = 60o
Therefore, sun’s elevation from the ground is 60o.
Example 2: An observer 1.5 m tall is 20.5 m away from a tower 22 m high. Determine the angle of elevation of the top of the tower from the eye of the observer.
Solution:
Let PQ be the height of the observer of 1.5 m.
Let AB be the height of the tower of 22 m.
And, let QB be the horizontal distance between the observer and the tower
PQ = MB = 1.5 m
Thus, AM = AB – MB
⇒ AM = 22 – 1.5 = 20.5Now, in ∆APM,
tan θ = AM / PM
⇒ tan θ = 20.5 / 20.5
⇒ tan θ = 1
⇒ θ = tan-1(1 )Hence, θ = 45o
Therefore, the angle of elevation of the top of the tower from the eye of the observer is 45o
Example 3: An airplane is flying h meters above the ground. At a particular instant, the angle of elevation of the plane from the eyes of a boy sitting on the ground is 60°. After some time, the angle of elevation changed to 30°. Find the distance covered by the plane during that time assuming it travelled in a straight line.
Solution:
Let x be the horizontal distance between the observer and plane at the first instant.
Let y be the horizontal distance between the observer and plane at the second instant.
And, BA = CD = h, Now in ∆OAB,
tan 60° = AB / OA
√3 = h / x
x = h / √3In ∆ OCD,
tan 30° = CD / OD
1/√3 = h / (x+y)
x + y = √3hDistance travelled by plane = AD = y
(x + y) − x = √3h − h / √3
y = (2 / √3)hSo, if the airplane is flying h meters above the ground, it would travel for (2/√3) h meters as the angle of elevation changes from 60° to 30°.
Example 4: From the top of the tower 30 m height a man is observing the base of a tree at an angle of depression measuring 30 degrees. Find the distance between the tree and the tower.
Solution:
In the above diagram AB represents the height of the tower, BC represents the distance between the foot of the tower and the foot of the tree.
Now we need to find the distance between the foot of the tower and the foot of the tree (BC). For that as angle of depression is given so by vertically opposite angle property of triangle ∠CAD = ∠BCA
In ∆BCA,
tan θ = Opposite side / Adjacent side
⇒ tan 30° = AB / BC
⇒ 1/√3 = 30 / BC
⇒ BC = 30√3
⇒ BC = 30 (1.732) [Approximately]
⇒ BC = 51.96 mSo, the distance between the tree and the tower is 51.96 m.
Example 5: From the top of a building 30 m high, the top and bottom of a tower are observed to have angles of depression 30° and 45° respectively. Find the height of the tower.
Solution:
Let AB be the building and CD be the tower.
The angle of depressions is given 30° and 45° to the top and bottom of the tower. So by vertically opposite triangle property ∠FBD = ∠EDB and ∠FBC = ∠ACB.
Now, AB = 30 m. Let DC = x.
Draw DE perpendicular AB. Then AE = CD = x.
Therefore BE = (30 – x) m.
In ∆ACB,
cot θ = Adjacent side / Opposite side
⇒ cot θ = AC / AB
⇒ cot 45° = AC / 30
⇒ AC = 30 [cot 45° = 1]So, DE = AC = 30 m
In ∆EDB,
tan θ = Opposite side / Adjacent side
⇒ tan 30° = BE / DE
⇒ 1/√3 = BE / 30
⇒ BE = 30 / √3Thus, CD = AE = AB – BE = 30 – (30 / √3)
⇒ CD = 30[1 – (1 / √3) ] mHeight of the tower is 30[1 – (1 / √3) ] m
FAQs on Height and Distances
What is the angle of depression in trigonometry?
The angle of depression is defined as the angle between the observer’s horizontal line of sight and the object when the observer is looking at the object which is placed downward.
What is the angle of elevation in trigonometry?
The angle of elevation is defined as the angle between the observer’s horizontal line of sight and the object when the observer is looking at the object which is placed upward.
Is the angle of elevation equal to depression?
Angle of elevation is always equal to angle of depression for any particular pair observer and object.
What is the line of sight in trigonometry?
Line of sight is the line drawn from the eye of observer to the point where the object is viewed by the observer.
Heights and Distances Practice Problems
- From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge is at a height of 3 m from the banks, find the width of the river.
- The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
- From a balloon vertically above a straight road, the angles of depression of two cars at an instant are found to be 45° and 60°. If the cars are 100 m apart, find the height of the balloon.
- The lower window of a house is at a height of 2 m above the ground and its upper window is 4 m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be 60° and 30°, respectively. Find the height of the balloon above the ground.
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