CH – 9 SOME APPLICATIONS OF TRIGONOMETRY
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Master real-world problem-solving, sketch precise diagrams, and score maximum marks in your CBSE Class 10 Board Exam!
Chapter Overview
In this chapter, you will learn how to apply basic trigonometric ratios to solve practical, real-world measurement problems:
- Using trigonometry to measure heights and distances in real-life situations without direct measurement.
- Understanding height and distance problems by converting verbal statements into geometric figures.
- Learning the exact concept of the angle of elevation when looking upward.
- Learning the exact concept of the angle of depression when looking downward.
- Drawing accurate, step-by-step geometric diagrams from given text.
- Choosing the correct trigonometric ratio (sin, cos, or tan) based on known and unknown sides.
- Finding unknown heights of towers, trees, or buildings, and distances between objects.
🎯 Board Exam Focus
Practise the following core skills carefully to ensure complete preparation for your board exam:
- Understanding the exact situation described in the problem word by word.
- Drawing a neat, correctly labelled diagram before starting any mathematical calculation.
- Identifying right-angled triangles clearly in both simple and complex figures.
- Applying the angle of elevation correctly from the horizontal line of sight.
- Applying the angle of depression correctly from the top horizontal reference line.
- Identifying the opposite side, adjacent side, and hypotenuse with respect to the reference angle.
- Selecting the appropriate trigonometric ratio (usually tan, sin, or cos) that connects the given side to the required side.
- Solving single-triangle and multi-triangle height and distance equations accurately.
Most Important Concepts
What Are Applications of Trigonometry?
In everyday life, we often need to know the height of a very tall object (like a mountain, a tower, or a tall building) or the distance across a wide river. It is impossible or very difficult to measure these directly using a tape or ruler.
Trigonometry helps us calculate these unknown heights and distances easily by measuring a distance on the ground and an angle formed by looking at the object.
Simple Real-Life Examples:
- Finding the height of a tall building standing on a level street.
- Finding the height of a transmission tower or a wind turbine.
- Finding the distance of a ship from a lighthouse standing at the edge of the sea.
Angle of Elevation
The angle of elevation is the angle formed between the horizontal line and the line of sight when an observer looks upward at an object placed above eye level.
When you raise your head to view something higher than you, your line of sight goes upward. The angle formed between this upward line of sight and the straight horizontal line from your eye is the angle of elevation.
Real-Life Example:
A person standing on the ground looking up at a kite flying high in the sky. The angle formed between the person’s horizontal line of vision and the string leading to the kite is the angle of elevation.
Angle of Depression
The angle of depression is the angle formed between the horizontal line and the line of sight when an observer looks downward at an object placed below eye level.
When you lower your head to view something lower than you, your line of sight goes downward. The angle formed between the horizontal line drawn at your eye level and the line leading down to the object is the angle of depression.
Real-Life Example:
A girl standing on a high balcony looking down at a ball lying on the street.
Key Distinction:
- Looking UPWARD → Angle of Elevation
- Looking DOWNWARD → Angle of Depression
🧠 Easy Way to Remember Elevation and Depression
Use this simple memory trick:
- ELEVATION = ELEVATOR going UP ⬆️ (Look UP)
- DEPRESSION = DEPRESSING / feeling DOWN ⬇️ (Look DOWN)
Understanding the Diagram
Drawing an accurate geometric diagram is the most critical step in solving any height-and-distance question. If your diagram is correct, half of your solution is already complete!
Follow these simple steps to draw a perfect diagram every time:
- Read carefully: Read the entire question once or twice to understand who is looking and what object is being observed.
- Mark the ground level: Draw a flat horizontal line at the bottom to represent the ground or water surface.
- Draw vertical objects: Draw straight vertical lines perpendicular (90°) to the ground to represent towers, poles, trees, or buildings.
- Locate the observer and object: Mark the exact position of the observer’s eye and the object.
- Draw the horizontal line: Always draw a dotted horizontal reference line extending straight out from the observer’s eye level.
- Draw the line of sight: Draw a straight line from the observer’s eye to the point being viewed on the object.
- Mark the angle: Place the given angle of elevation or depression directly between the line of sight and the horizontal line.
- Identify the right-angled triangle: Identify the 90° angle formed between the vertical height and the horizontal ground, forming your working right triangle.
Finding the Correct Sides
Once your right-angled triangle is ready, label its three sides according to the reference angle (θ) you are using:
- Hypotenuse (Hyp): The longest side of the triangle, located directly opposite to the 90° right angle.
- Opposite Side (Opp) / Perpendicular: The side situated directly opposite (facing) your reference angle θ.
- Adjacent Side (Adj) / Base: The horizontal or vertical side that touches your reference angle θ (alongside the hypotenuse).
Remember: The hypotenuse is fixed facing the 90° angle. The opposite and adjacent sides change depending on which angle inside the triangle you choose as your reference angle.
Choosing the Correct Trigonometric Ratio
To select the right ratio, identify which side value you know and which side value you need to find.
- Use TAN (tan θ = Opposite ÷ Adjacent):Used in over 80% of height-and-distance questions! Choose tan when you are dealing with the vertical height (Opposite) and the horizontal ground distance (Adjacent), and the hypotenuse is not needed.
- Use SIN (sin θ = Opposite ÷ Hypotenuse):Choose sin when your problem involves the vertical height (Opposite) and the length of a string, rope, or ladder (Hypotenuse).
- Use COS (cos θ = Adjacent ÷ Hypotenuse):Choose cos when your problem involves the horizontal ground distance (Adjacent) and the slanted length/hypotenuse.
Solving a Height and Distance Problem
Follow this standard procedure to solve any problem step by step:
- Draw a clean, labelled diagram.
- Write down all given values clearly (e.g., angle θ, given side lengths).
- State the right-angled triangle you are considering.
- State the chosen trigonometric ratio.
- Substitute the known numerical values into the equation.
- Perform algebraic simplification and step-by-step calculation.
- Write the final answer clearly with the correct unit (e.g., metres or centimetres).
Step-by-Step Worked Example
Question:
A vertical pole stands on level ground. From a point on the ground which is 20 metres away from the foot of the pole, the angle of elevation of the top of the pole is 30°. Find the height of the pole.
A (Top of pole)
|\
| \
| \
| \ Line of sight
Height of | \
pole (h) | \
| \
|___30°_\
B C
<--20m-->
Foot of pole
Step-by-step Solution:
- Step 1: Identify given informationLet AB be the height of the vertical pole, denoted by h metres.Let C be the point on the ground 20 metres away from the foot of the pole B.Distance BC = 20 m.Angle of elevation ∠ACB = 30°.Angle ∠ABC = 90° (since the pole is vertical).
- Step 2: Identify sides and choose ratioFor reference angle 30° at C:Opposite side = AB = hAdjacent side = BC = 20 mSince we know the Adjacent side and want to find the Opposite side, we use the tan ratio.
- Step 3: Apply the ratio equationIn right-angled triangle ABC:tan 30° = Opposite ÷ Adjacenttan 30° = AB ÷ BC
- Step 4: Substitute values1 ÷ √3 = h ÷ 20
- Step 5: Calculate hh = 20 ÷ √3
Multiply numerator and denominator by √3 to rationalize the denominator:h = (20 × √3) ÷ (√3 × √3)h = (20√3) ÷ 3 metres - Step 6: Final StatementThe height of the pole is (20√3)/3 metres (or 11.55 m approximately if √3 = 1.732 is used).
Problems Involving Angle of Depression
When a question describes an observer looking down from a height (such as a lighthouse, building, or cliff), the angle given is an angle of depression.
Understanding Alternate Interior Angles
Because the observer’s horizontal line of sight is parallel to the ground horizontal line, the angle of depression formed at the top is equal to the angle of elevation formed at the ground object.
This is due to the geometric rule of alternate interior angles formed by a transversal line cutting through parallel horizontal lines.
Horizontal Line at Top P •-------------• Q (Parallel to Ground)
\ 30°
\
\ Line of Sight
\
\ 30°
Ground Level A •----• B
Ground Object
- Line PQ is parallel to Line AB.
- Line PB acts as a transversal line.
- Angle QPB (Angle of Depression) = Angle PBA (Angle of Elevation at ground level).
Step-by-Step Worked Example (Angle of Depression)
Question:
From the top of a 50 metre high cliff, the angle of depression of a boat on the sea is 45°. Find the distance of the boat from the foot of the cliff.
Observer's Line P •-----------• Q (Horizontal)
\ 45°
\
\ Line of sight
\
Cliff \
(50 m) \ 45°
•-----------•
A B (Boat)
<--- x --->
Step-by-step Solution:
- Step 1: Identify given informationLet PA be the height of the cliff = 50 m.Let B be the position of the boat on the sea surface.Let PQ be the horizontal line at eye level, parallel to ground AB.Angle of depression ∠QPB = 45°.
- Step 2: Convert to angle inside the triangleSince horizontal line PQ is parallel to sea surface AB, by alternate interior angles:∠PBA = ∠QPB = 45°
- Step 3: Apply trigonometric ratioIn right-angled triangle PAB (right-angled at A):tan 45° = Opposite ÷ Adjacenttan 45° = PA ÷ AB
- Step 4: Substitute values1 = 50 ÷ ABAB = 50 metres
- Step 5: Final StatementThe distance of the boat from the foot of the cliff is 50 metres.
📌 Important Concepts & Results
- Horizontal Line: A straight line parallel to the ground surface drawn straight out from the observer’s eye level.
- Line of Sight: The imaginary straight line drawn directly from the eye of an observer to the point being viewed on an object.
- Angle of Elevation: The angle formed between the horizontal line and the line of sight when looking upward at an object above eye level.
- Angle of Depression: The angle formed between the horizontal line and the line of sight when looking downward at an object below eye level.
- Equality of Angles: The angle of depression from an observer to a ground point equals the angle of elevation from the ground point back to the observer (by alternate interior angles of parallel horizontal lines).
- Right-Angled Triangle: Triangle with one 90° angle formed between vertical heights and horizontal ground levels.
- Primary Trigonometric Ratios:
- tan θ = Opposite ÷ Adjacent (most commonly used)
- sin θ = Opposite ÷ Hypotenuse
- cos θ = Adjacent ÷ Hypotenuse
🧠 How to Know What to Do in a Question?
Follow this simple decision guide when solving questions:
- If the question asks for height: Look for the vertical side (Opposite). If horizontal distance is given, use tan θ. If string/rope/ladder length is given, use sin θ.
- If the question asks for distance: Look for the horizontal side (Adjacent). If height is given, use tan θ. If slanted length is given, use cos θ.
- If the observer looks upward: Measure the angle of elevation from the ground horizontal line looking up.
- If the observer looks downward: Draw a horizontal dotted line at the top observer position, mark the angle of depression, and transfer it inside the ground triangle using alternate interior angles.
- If two angles are given (two triangles): Write tan equations for both right-angled triangles separately, and solve the algebraic equations by substitution or elimination.
🧠 Concepts Students Find Difficult
- Drawing accurate diagrams: Converting lengthy English text into a simple geometric figure.
- Locating line of sight: Understanding that line of sight connects the observer’s eye to the exact target point.
- Confusing elevation and depression: Placing the angle of depression inside the triangle next to the vertical wall instead of between the line of sight and the top horizontal line.
- Observer height inclusion: Forgetting to subtract or add the observer’s height when observer height is specified (e.g., 1.5 m tall boy).
- Identifying reference angle sides: Switching opposite and adjacent sides by mistake when changing reference angles.
- Choosing incorrect ratios: Attempting to use tan when string or ladder length (hypotenuse) is asked.
- Omitting units: Writing final values as plain numbers without adding ‘m’ or ‘cm’.
⚠️ Common Mistakes Students Make
- Drawing angle of depression against the vertical wall:
- Mistake: Placing the angle of depression between the vertical building wall and the line of sight.
- Fix: Always draw a dotted horizontal line at the top eye level first. Place the angle between that horizontal line and the line of sight.
- Ignoring observer height:
- Mistake: Treating a 1.5 m tall observer as a point on the ground, leading to height calculation errors.
- Fix: If observer height is given, draw a rectangle at the base. Calculate triangle height first, then add observer height at the end.
- Confusing string length with ground distance:
- Mistake: Treating the string of a kite or a ladder as the adjacent ground distance.
- Fix: Remember that ladders, kite strings, and ramp surfaces are slanted lines forming the hypotenuse.
- Selecting tan instead of sin for ladder/rope questions:
- Mistake: Using tan θ when the question asks for ladder or rope length.
- Fix: When hypotenuse is required and opposite height is known, always select sin θ = Opposite ÷ Hypotenuse.
- Using incorrect standard values:
- Mistake: Writing tan 30° = √3 or tan 60° = 1/√3.
- Fix: Double-check standard values before solving: tan 30° = 1/√3, tan 45° = 1, tan 60° = √3.
- Forgetting to rationalize denominators:
- Mistake: Leaving answers like 15 ÷ √3 m unfinished.
- Fix: Multiply top and bottom by √3 to get clean simplified forms like 5√3 m.
- Forgetting units in final answer:
- Mistake: Writing “Height of tower = 25” without units.
- Fix: Always write units clearly, such as “Height of tower = 25 m”.
🔥 Priority Revision
⭐⭐⭐ HIGH-PRIORITY PRACTICE
- Double-triangle problems involving two angles of elevation (e.g., 30° and 60°) from different ground points.
- Double-triangle problems involving angles of depression from the top of a tower to two objects on the ground.
- Combination problems where a tower or flagstaff is mounted on top of a building or hill.
⭐⭐ IMPORTANT PRACTICE
- Questions where an observer has a specific height (e.g., 1.5 m observer looking at a chimney or building).
- Kite string or ladder length calculations involving sin θ or cos θ.
- River width or road width problems between two opposite points.
⭐ ADDITIONAL REVISION
- Simple single-triangle direct height calculations using tan 30°, tan 45°, or tan 60°.
- Speed, distance, and time problems involving changing angles of depression over time.
⚡ Some Applications of Trigonometry Quick Revision Sheet
- Line of Sight: Line connecting observer’s eye to target point.
- Angle of Elevation: Angle above horizontal line when looking UP.
- Angle of Depression: Angle below horizontal line when looking DOWN.
- Angle Rule: Angle of depression at top = Angle of elevation at ground (alternate interior angles).
- Diagram Steps:
- Horizontal ground line.
- Vertical height line (90°).
- Slanted line of sight.
- Label given angles and lengths.
- Selecting Ratios:
- Height & Ground Distance → tan θ
- Height & Slanted Length (Ladder/Rope/String) → sin θ
- Ground Distance & Slanted Length → cos θ
- Standard Key Values:
- tan 30° = 1/√3, tan 45° = 1, tan 60° = √3
- sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2
- cos 30° = √3/2, cos 45° = 1/√2, cos 60° = 1/2
- Important Reminder: Always include measurement units (m, cm) in your final line!
📝 Test Yourself
- A vertical tower stands on level ground. From a point on the ground 30 metres away from the foot of the tower, the angle of elevation of its top is 30°. Find the height of the tower.
- A ladder leaning against a vertical wall makes an angle of 60° with the ground. If the foot of the ladder is 4 metres away from the wall, find the length of the ladder.
- A kite is flying at a height of 75 metres above ground level. The string attached to the kite is temporarily tied to a point on the ground. Find the length of the string if its inclination with the ground is 45°.
- An observer 1.6 metres tall stands 28.4 metres away from a vertical tower. The angle of elevation of the top of the tower from the observer’s eyes is 45°. Determine the total height of the tower.
- From the top of a 60 metre high building, the angle of depression of a car parked on the ground is 30°. Calculate the distance of the car from the base of the building.
- A flagstaff stands vertically on top of a 15 metre high building. From a point on the ground, the angles of elevation of the bottom and top of the flagstaff are 45° and 60° respectively. Find the height of the flagstaff.
- The shadow of a vertical tower on level ground increases by 30 metres when the Sun’s altitude changes from 60° to 30°. Find the height of the tower.
- Two vertical poles of equal heights stand on opposite sides of a 100 metre wide road. From a point on the road between them, the angles of elevation of their tops are 60° and 30°. Find the height of each pole and the position of the point.
- From the top of a 7 metre high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the total height of the cable tower.
- 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°. If the bridge is at a height of 4 metres above the banks, calculate the total width of the river.
🎯 Before Your Exam
☐ I fully understand the concept of angle of elevation. ☐ I fully understand the concept of angle of depression. ☐ I can convert any problem statement into an accurate diagram. ☐ I can correctly locate the right-angled triangle in a figure. ☐ I can correctly label opposite, adjacent, and hypotenuse sides. ☐ I can select the correct trigonometric ratio (sin, cos, tan) instantly. ☐ I know how to convert angle of depression to alternate interior angle. ☐ I can solve double-triangle height-and-distance equations step by step. ☐ I remember to add observer height when mentioned in the question. ☐ I always write proper units (metres, cm) in my final answer.
Mastering diagrams and ratios will make every height-and-distance problem simple and easy to score full marks in your exam!