Some Applications of Trigonometry
Chapter 9 | Mathematics
Complete NCERT notes, line of sight, angle of elevation, angle of depression, heights and distances problems, and solved examples for CBSE Class 10 Board Exam 2025-26.
π Introduction
Trigonometry is widely used in real-life applications to measure heights and distances indirectly. It is used in navigation, surveying, astronomy, architecture, and engineering.
π‘ Theodolite is an instrument used to measure angles of elevation and depression.
π Angle of Elevation
π Definition
When we look up at an object, the angle formed between the line of sight and the horizontal line is called the angle of elevation.
π Angle of Depression
π Definition
When we look down at an object, the angle formed between the line of sight and the horizontal line is called the angle of depression.
π‘ Angle of elevation from one point = Angle of depression from the other point when both points are at different heights.
π NCERT Solved Examples
Example 1: Angle of Elevation
A tower stands vertically on the ground. From a point on the ground 15 m away from the foot of the tower, the angle of elevation of the top of the tower is 60Β°. Find the height of the tower.
tan 60Β° = Height / 15 β β3 = h/15 β h = 15β3 = 15 Γ 1.732 = 25.98 m
β΄ Height of tower = 15β3 m (β 25.98 m)
Example 2: Angle of Depression
From the top of a 7 m high building, the angle of depression of the foot of a tower is 45Β°. Find the distance between the building and the tower.
tan 45Β° = Height of building / Distance β 1 = 7 / d β d = 7 m
β΄ Distance = 7 m
Example 3: Two Angles of Elevation
The angle of elevation of the top of a tower from a point on the ground is 30Β°. After moving 50 m towards the tower, the angle becomes 60Β°. Find the height of the tower.
Let height = h. From first point: tan30Β° = h/x β 1/β3 = h/x β x = hβ3
From second point: tan60Β° = h/(x-50) β β3 = h/(x-50) β x-50 = h/β3
Substitute: hβ3 - 50 = h/β3 β Multiply by β3: 3h - 50β3 = h β 2h = 50β3 β h = 25β3 m
Example 4: Two Objects on Same Side
From the top of a 60 m high lighthouse, the angles of depression of two ships on the same side are 30Β° and 45Β°. Find the distance between the ships.
tan45Β° = 60/dβ β 1 = 60/dβ β dβ = 60 m
tan30Β° = 60/dβ β 1/β3 = 60/dβ β dβ = 60β3 m
Distance between ships = dβ - dβ = 60β3 - 60 = 60(β3 - 1) m β 60 Γ 0.732 = 43.92 m
Example 5: Airplane Problem
An airplane flying at a height of 3000 m passes vertically above another airplane at an instant when the angles of elevation of the two airplanes from the same point on the ground are 60Β° and 45Β° respectively. Find the vertical distance between them.
Let distance of point from ground = d. For lower plane: tan45Β° = hβ/d β 1 = hβ/d β hβ = d
For higher plane: tan60Β° = hβ/d β β3 = hβ/d β hβ = dβ3
But hβ = 3000 m β d = 3000/β3 = 1000β3 m
Then hβ = 1000β3 m
Vertical distance = hβ - hβ = 3000 - 1000β3 = 1000(3 - β3) m
β Key Points for Board Exam
- Angle of Elevation: Measured upward from horizontal.
- Angle of Depression: Measured downward from horizontal.
- Angle of elevation = Angle of depression when looking from one point to another at different heights.
- Always draw a clear figure showing the horizontal line, line of sight, and known distances.
- Use tan ratio most frequently for height-distance problems.
- When an object moves towards/away from the observer, the angle of elevation changes.
- For two points on the same side of the observer, subtract distances.
- For two points on opposite sides, add distances.
- Memorize values of tan for 0Β°, 30Β°, 45Β°, 60Β°, 90Β°.
- Rationalize denominators in final answers.