CBSE Class 10 Mathematics Applications of Trigonometry Notes
About This Chapter
Applications of Trigonometry is Chapter 9 of the CBSE Class 10 Mathematics syllabus. This chapter builds on the foundational concepts of trigonometric ratios introduced in the previous chapter and applies them to solve real-world problems involving heights and distances. Students learn to calculate inaccessible measurements using angles of elevation and depression along with basic trigonometric identities.
In everyday life, trigonometry is used by engineers, architects, surveyors, and navigators. From finding the height of a mountain or a lighthouse to determining the distance of a ship from the shore, this chapter provides the mathematical tools that professionals rely on. Understanding these applications makes abstract concepts concrete and highly practical.
In the CBSE Board Examination, this chapter typically carries 6 to 8 marks out of 80, making it a moderate-weightage chapter. Questions usually appear as 2-mark or 3-mark short answer questions and occasionally as 4-mark long answer questions. Mastering this chapter helps students secure full marks on predictable question patterns.
These notes cover every concept, formula, and example type you need for your board exam preparation. A comprehensive PDF version is attached below for download and offline study.
What You Will Learn:
• Meaning and definition of angle of elevation and angle of depression
• How to identify and draw the correct right-angled triangle from a word problem
• Application of sin, cos, tan and their standard angle values (30, 45, 60 degrees)
• Solving multi-step problems involving two observers or two angles
• Identifying common errors and applying exam-focused strategies for full marks
The complete PDF of these notes is attached below for download.
1. Introduction and Definition
Trigonometry is the branch of mathematics that studies the relationships between angles and sides of triangles. While the previous chapter introduced trigonometric ratios, this chapter focuses on applying those ratios to practical problems involving heights and distances that cannot be measured directly.
1.1 What Are Applications of Trigonometry?
In real-world scenarios, we often need to find the height of a building, the distance of a ship from a cliff, or the width of a river, without physically measuring them. By standing at a point and measuring the angle to the top or bottom of the object, and knowing one measurable distance, we can calculate the unknown using trigonometric ratios.
1.2 Key Terminology
Term | Definition | Diagram Reference |
Angle of Elevation | Angle formed between the horizontal line and the line of sight when looking upward at an object | Observer looks up toward top of tower |
Angle of Depression | Angle formed between the horizontal line and the line of sight when looking downward at an object | Observer looks down toward base of object below |
Line of Sight | The straight line from the observer's eye to the object being viewed | Hypotenuse of the right triangle |
Horizontal Level | The horizontal line through the observer's eye, parallel to the ground | Base reference line |
Foot of the Tower | The base of a vertical structure on the ground level | Adjacent side reference point |
2. Key Concepts and Components
2.1 Angle of Elevation
When an observer at point O looks upward at an object P at the top of a tower, the angle formed between the horizontal line OA and the line of sight OP is called the angle of elevation. This angle is always measured from the horizontal upward.
Angle of Elevation Setup |
tan(theta) = Opposite / Adjacent = Height of Tower / Horizontal Distance |
2.2 Angle of Depression
When an observer standing at a height looks downward at an object B below, the angle formed between the horizontal line and the line of sight toward the object is called the angle of depression. By alternate interior angles, the angle of depression from the top equals the angle of elevation from the bottom.
Key Property |
Angle of Depression (from top) = Angle of Elevation (from bottom) [Alternate Interior Angles] |
2.3 Standard Trigonometric Values Used
Angle | sin | cos | tan |
0 degrees | 0 | 1 | 0 |
30 degrees | 1/2 | sqrt(3)/2 | 1/sqrt(3) |
45 degrees | 1/sqrt(2) | 1/sqrt(2) | 1 |
60 degrees | sqrt(3)/2 | 1/2 | sqrt(3) |
90 degrees | 1 | 0 | Undefined |
3. Core Formulas with Derivations
3.1 Basic Trigonometric Ratio Formulas
In a right-angled triangle with angle theta, opposite side (O), adjacent side (A), and hypotenuse (H):
Primary Ratios |
sin(theta) = O/H cos(theta) = A/H tan(theta) = O/A |
Reciprocal Ratios |
cosec(theta) = H/O sec(theta) = H/A cot(theta) = A/O |
3.2 Height and Distance Formulas
From these, the most commonly used formulas in height and distance problems are:
Height from Angle of Elevation |
Height (h) = Distance (d) x tan(theta) |
Distance from Height |
Horizontal Distance (d) = Height (h) / tan(theta) = h x cot(theta) |
3.3 Two-Angle Problems
When an observer views a tower from two different positions, or two observers view the same object, the problem involves two angles. The Pythagoras theorem and simultaneous equation approach are used.
Two-Angle Formula |
h = d x tan(alpha) AND h = (d + x) x tan(beta), where x = additional distance |
4. Solved Examples
Example 1: Finding Height of a Tower
Problem: A tower stands vertically on the ground. From a point on the ground 30 m away from the foot of the tower, the angle of elevation of the top of the tower is 60 degrees. Find the height of the tower.
Solution: Let the height of the tower = h Horizontal distance = 30 m Angle of elevation = 60 degrees
Using: tan(60 degrees) = h / 30 => sqrt(3) = h / 30 => h = 30 x sqrt(3) => h = 30 x 1.732 = 51.96 m
Therefore, the height of the tower is 30*sqrt(3) m, approximately 51.96 m. |
Example 2: Angle of Depression
Problem: From the top of a cliff 100 m high, the angle of depression of a ship at sea is 30 degrees. Find the horizontal distance of the ship from the base of the cliff.
Solution: Height of cliff (h) = 100 m Angle of depression = 30 degrees By alternate interior angles, angle of elevation from ship = 30 degrees
tan(30 degrees) = 100 / d => 1/sqrt(3) = 100 / d => d = 100 x sqrt(3) = 100*sqrt(3) m
The ship is 100*sqrt(3) m (approximately 173.2 m) from the base of the cliff. |
Example 3: Two Observers / Two Angles
Problem: The angles of elevation of the top of a tower from two points P and Q at distances 4 m and 9 m from the base, on the same side, are complementary. Prove that the height of the tower is 6 m.
Solution: Let angle of elevation from P = theta, from Q = (90 - theta) [complementary] Height of tower = h
From P: tan(theta) = h/4 => h = 4*tan(theta) ...(1) From Q: tan(90 - theta) = h/9 => cot(theta) = h/9 ...(2)
Multiplying (1) and (2): h x h = 4*tan(theta) x 9*cot(theta) h^2 = 36 x (tan(theta) x cot(theta)) h^2 = 36 x 1 = 36 h = 6 m [Proved] |
Example 4: Moving Observer
Problem: A man standing on top of a 20 m building observes that the angle of depression of a car changes from 30 to 60 degrees as the car moves toward the building. Find the distance traveled by the car.
Solution: Height of building = 20 m Initial angle of depression = 30 deg, Final = 60 deg
Initial distance: tan(30) = 20/d1 => d1 = 20*sqrt(3) Final distance: tan(60) = 20/d2 => d2 = 20/sqrt(3) = 20*sqrt(3)/3
Distance traveled = d1 - d2 = 20*sqrt(3) - 20*sqrt(3)/3 = 20*sqrt(3) x (1 - 1/3) = 20*sqrt(3) x 2/3 = 40*sqrt(3)/3 m |
Example 5: Width of River
Problem: A boy standing on one bank of a river observes the angle of elevation of a tree on the opposite bank to be 60 degrees. Moving 20 m away, the angle becomes 30 degrees. Find the width of the river.
Solution: Let width of river = w, height of tree = h
From first point: tan(60) = h/w => h = w*sqrt(3) ...(1) From second point: tan(30) = h/(w+20) => h = (w+20)/sqrt(3) ...(2)
From (1) and (2): w*sqrt(3) = (w+20)/sqrt(3) 3w = w + 20 2w = 20 => w = 10 m
Width of river = 10 m |
5. Applications and Special Cases
5.1 Real-World Applications
• Navigation: Sailors use angles of elevation and depression to determine distances from lighthouses and cliffs.
• Architecture: Engineers use trigonometry to calculate the safe height and angle of incline for ramps and staircases.
• Aviation: Pilots use angles of descent (angle of depression) during approach and landing.
• Surveying: Land surveyors use theodolites to measure angles and compute heights of inaccessible terrain.
• Astronomy: The height and distance of stars and planets are estimated using parallax angles based on trigonometric principles.
5.2 Special Case: Complementary Angles
When two angles of elevation are complementary (sum = 90 degrees), their product of tangents equals 1. This gives:
Complementary Angles Property |
h^2 = d1 x d2 (where d1 and d2 are the two horizontal distances) |
5.3 Special Case: 45-Degree Angle
When the angle of elevation is 45 degrees, tan(45) = 1. This means the height of the tower equals the horizontal distance from the observer. This is a very common scenario in board exam problems.
6. Formula Summary Table
The table below summarizes all key formulas used in this chapter:
Formula / Concept | Expression | When to Use |
Height from angle | h = d x tan(theta) | When distance is known, find height |
Distance from angle | d = h / tan(theta) = h x cot(theta) | When height is known, find distance |
Hypotenuse (line of sight) | L = h / sin(theta) | When height and angle known |
Complementary angles | h^2 = d1 x d2 | When two angles sum to 90 deg |
tan(30) | 1/sqrt(3) = sqrt(3)/3 | Standard value |
tan(45) | 1 | Height equals distance |
tan(60) | sqrt(3) | Standard value |
Angle of depression = Elevation | Both angles are equal | Alternate interior angles property |
7. Key Theorems and Properties
7.1 Alternate Interior Angles Theorem
When a horizontal line is cut by a transversal (line of sight), the angle of depression from the upper point equals the angle of elevation from the lower point. This is a consequence of the Alternate Interior Angles Theorem from geometry and is used in almost every height-and-distance problem.
7.2 Pythagoras Theorem in Trigonometry Problems
In any right-angled triangle formed in a height-and-distance problem, the Pythagoras Theorem (H^2 = P^2 + B^2) can be used to find the third side when two sides are known. This often comes into play when the line of sight (hypotenuse) needs to be calculated.
Pythagoras Theorem |
H^2 = P^2 + B^2 (where H = hypotenuse, P = perpendicular, B = base) |
7.3 Trigonometric Identities Used
Key Identity 1 |
sin^2(theta) + cos^2(theta) = 1 |
Key Identity 2 |
tan(theta) = sin(theta) / cos(theta) |
Key Identity 3 |
cot(theta) = cos(theta) / sin(theta) = 1 / tan(theta) |
8. Common Mistakes and Exam Tips
8.1 Common Mistakes to Avoid
1. Confusing angle of elevation with angle of depression: Always check the observer's position. If the observer is below the object, it is elevation; if above, it is depression.
2. Not drawing a diagram: Most errors arise from visualizing the problem incorrectly. Always draw the right triangle before writing equations.
3. Using wrong trigonometric ratio: Identify which two sides are known/required (opposite, adjacent, hypotenuse) and choose sin, cos, or tan accordingly.
4. Forgetting alternate interior angles: The angle of depression equals the angle of elevation only when you correctly apply the alternate angles property.
5. Rounding errors: Always keep answers in surd (sqrt) form unless the question asks for a decimal approximation.
8.2 Exam Tips for Full Marks
• Always label the diagram with all given values including angles, heights, and distances before starting calculations.
• Show each step clearly. Even if the final answer is wrong due to a calculation error, you earn marks for method.
• Memorize standard values of sin, cos, and tan for 0, 30, 45, 60, and 90 degrees.
• For two-angle problems, set up two equations and solve simultaneously. This approach is almost always tested.
• Check your answer by substituting back: if you found h, verify that tan(theta) = h/d gives the angle stated in the problem.
9. Practice Questions
Test your understanding with the following questions arranged by marks:
Category A: 1 Mark Questions (MCQ / Very Short Answer)
If the angle of elevation of the sun is 45 degrees, find the length of the shadow of a pole 10 m high.
What is the value of tan(60 degrees)?
A pole 6 m high casts a shadow 6*sqrt(3) m long. What is the angle of elevation of the sun?
If angle of depression of a point from a 50 m cliff is 30 degrees, the horizontal distance is ___ m.
The angle of elevation equals the angle of depression. True or False? Justify.
A ladder 10 m long makes an angle of 60 degrees with the ground. What is the height it reaches on the wall?
Category B: 3 Mark Questions (Short Answer)
From a point on the ground, the angle of elevation of a tower is 30 degrees. On walking 20 m toward the tower, the angle becomes 60 degrees. Find the height of the tower.
Two poles are 80 m apart and on the same side. The angles of elevation of the top of the taller pole from the foot of the shorter are 60 degrees and 30 degrees respectively. Find the height of each pole.
A man on the top of a vertical tower observes a car moving at a uniform speed coming directly toward the tower. He takes 18 minutes to find that the angle of depression changed from 30 to 60 degrees. How soon will the car reach the base?
A kite is flying at a height of 75 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60 degrees. Find the length of the string, assuming no slack.
From the top of a 7 m building, the angle of elevation of a cable tower is 60 degrees and the angle of depression of its foot is 45 degrees. Find the height of the tower.
Category C: 5 Mark Questions (Long Answer)
The angle of elevation of the top of a hill from the foot of a tower is 60 degrees and the angle of elevation of the top of the tower from the foot of the hill is 30 degrees. If the tower is 50 m high, find the height of the hill.
From the top of a lighthouse 60 m high, the angles of depression of two ships due east of the lighthouse are 30 degrees and 45 degrees. Find the distance between the ships.
Two boats are sailing toward a cliff of height 150 m from opposite sides. The angles of elevation of the top of the cliff from the two boats are 30 degrees and 45 degrees respectively. Find the distance between the two boats.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
The angle of elevation of the top of a vertical tower PQ from a point X is 60 degrees. At a point Y, 40 m vertically above X, the angle of elevation of P is 45 degrees. Find the height of the tower PQ and the distance XQ.
CBSE Class 10 Syllabus |
CBSE Class 10 Notes |