The topic heights and distance is one of the applications of Trigonometry which is extensively used in real-life. In the height and distances application of trigonometry, the following concepts are included:
- Measuring the heights of towers or big mountains
- Determining the distance of the shore from the sea
- Finding the distance between two celestial bodies
It should be noted that finding the height of bodies and distances between two objects is one of the most important applications of trigonometry.
What are Heights and Distances?
- Definition 1: The line which is drawn from the eyes of the observer to the point being viewed on the object is known as the line of sight.
- Definition 2: The angle of elevation of the point on the object (above horizontal level) viewed by the observer is the angle which is formed by the line of sight with the horizontal level.
- Definition 3: The angle of depression of the point on the object (below horizontal level) viewed by the observer is the angle which is formed by the line of sight with the horizontal level.
How to Find Heights and Distances?
To measure heights and distances of different objects, we use trigonometric ratios. For example, in fig.1, there is a guy 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 angle of elevation. Similarly, in fig. 2, PQ is the line of sight, PR is the horizontal level and ∠P is called angle of depression.
The angles of elevation and depression are usually measured by a device called Inclinometer or Clinometer.
To solve height and distance problems, trigonometric ratios for standard angles (Table 1) should be remembered.
Table 1: Trigonometric Ratios of Standard Angles
|tan C||0||1/√3||1||√3||Not Defined|
|csc C||Not Defined||2||√2||2/√3||1|
|sec C||1||2/√3||√2||2||Not Defined|
|cot C||Not Defined||√3||1||1/√3||0|
Solve a question based on the above topic for better understanding of this topic.
Heights and Distances Example Question
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.
The scenario explained in the question can be drawn as shown in the figure.
In ∆ OAB,
tan60° = ABOA
√3 = h/x
x = h/√3
In ∆ OCD,
tan30° = 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 °.
More Topics Related to Height and Distances:
|Trigonometry||Some Applications Of Trigonometry Class 10 Notes|
|Angle of Elevation||Important Questions Class 10 Maths Chapter 9 Applications of Trigonometry|
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