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Andhra Pradesh (BIEAP)Class 10 Mathematics★ AP Board (BSEAP) SSC Mathematics · ~6 marks · Board: AP Board

Applications of Trigonometry — Heights and Distances

ANGLE OF ELEVATION (looking up) and ANGLE OF DEPRESSION (looking down). Solve real-world problems involving heights of buildings, towers, and distances using trigonometry. AP Board Class 10 Mathematics.

⏱ 8 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Distinguish between angle of elevation and angle of depression
2Solve single-observation problems: find height of tower, tree, building
3Solve two-observation problems: object viewed from two positions or from two different heights
4Use the key relation: angle of elevation from A to B = angle of depression from B to A
5Set up simultaneous equations for two-triangle problems

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Why this chapter matters

Applications of Trigonometry — Heights and Distances — is a GUARANTEED exam question in AP SSC. The chapter is entirely about using tan (and sometimes sin/cos) to find heights and distances in real-world scenarios. One-observation problems (single right triangle) earn 2 marks; two-observation problems earn 4 marks. Every problem follows the same approach: draw a diagram, identify the right triangle(s), apply tan, solve. Students who can draw clear diagrams consistently score full marks.

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Applications of Trigonometry — Heights and Distances

Key Definitions

Angle of Elevation: Angle formed by the LINE OF SIGHT with the HORIZONTAL when looking UP.
Angle of Depression: Angle formed when looking DOWN.

Problem Types

Single observation: One right triangle. Use tan θ = height/distance.
Two observations: Two positions from which the object is viewed. Two equations using tan. Solve simultaneously.

Solving Strategy

DRAW a clear diagram. Label everything. 2. Identify the RIGHT TRIANGLE(S). 3. Choose the appropriate trigonometric ratio (usually tan). 4. Solve.

Common Mistake: Confusing angle of elevation with angle of depression. 'Angle of elevation from A to B = Angle of depression from B to A. They are EQUAL (alternate interior angles).'

Worked Examples — Heights and Distances

Example 1 — Height of a Tower (Single Observation)

A tower casts a shadow of length 50 m when the sun's elevation is 30°. Find the height of the tower.

Step 1: Draw and label. Let AB be the tower (height h). BC is the shadow (50 m). The angle of elevation of the sun = ∠ACB = 30°.

Step 2: Identify the ratio. In right ΔABC (∠B = 90°): tan 30° = AB/BC = h/50.

Step 3: Solve. h = 50 × tan 30° = 50 × (1/√3) = 50/√3 = (50√3)/3 ≈ 28.87 m.

'Always rationalise the denominator. Leave the answer in exact form (50/√3 m) unless a decimal approximation is explicitly asked.'

Example 2 — Distance Using Angle of Depression

From the top of a building 60 m high, the angle of depression of a car on the ground is 45°. Find the distance of the car from the base of the building.

Step 1: Building AB (height 60 m). Car at C. Angle of depression from A to C = 45°.

Step 2: Angle of depression from A = Angle of elevation from C = ∠ACB = 45° (alternate interior angles).

Step 3: In right ΔABC: tan 45° = AB/BC = 60/BC → 1 = 60/BC → BC = 60 m.

'Angle of depression from top equals angle of elevation from bottom. They are alternate interior angles and are EQUAL.'

Example 3 — Observer Height Included

A man 1.5 m tall stands 20 m from a tower. He observes the angle of elevation of the top of the tower to be 60°. Find the height of the tower.

Step 1: Man's eye at E (1.5 m above ground). Tower AB. Horizontal distance DE = 20 m. ∠CED = 60°.

Step 2: In right ΔCED: tan 60° = CD/DE → √3 = CD/20 → CD = 20√3 ≈ 34.64 m.

Step 3: Height AB = CD + AD = 20√3 + 1.5 = 20√3 + 1.5 ≈ 36.14 m.

'Do NOT forget to ADD the observer's eye height. This is the most common mistake in such problems.'

Example 4 — Two Ships from Lighthouse

A lighthouse is 100 m high. From its top, the angles of depression of two ships on the same side are 30° and 60°. Find the distance between the ships.

Step 1: Lighthouse AB (100 m). Ships at C (farther) and D (closer). Angles of depression 30° and 60°.

Step 2: For closer ship D: tan 60° = 100/AD → AD = 100/√3 = (100√3)/3 m. For farther ship C: tan 30° = 100/AC → AC = 100√3 m.

Step 3: CD = AC − AD = 100√3 − 100/√3 = 100(√3 − 1/√3) = 100(2/√3) = 200/√3 = (200√3)/3 ≈ 115.47 m.

'When both angles of depression are from the SAME point, distance between objects = |distance₁ − distance₂|.'

Example 5 — Walking Towards a Building

A 2 m tall boy observes the angle of elevation of a building as 30°. After walking 30 m towards the building, the angle becomes 60°. Find the height of the building.

Step 1: Let CD = h be the building height above eye level (2 m). Initial distance = x. tan 30° = h/x → 1/√3 = h/x → x = h√3. After moving 30 m: tan 60° = h/(x−30) → √3 = h/(x−30).

Step 2: Substitute x: √3 = h/(h√3 − 30) → 3h − 30√3 = h → 2h = 30√3 → h = 15√3 ≈ 25.98 m.

Step 3: Total height = 15√3 + 2 ≈ 27.98 m.

'Two-position problems are VERY common in AP exams. Set up TWO equations using tan and solve simultaneously.'

Example 6 — Flagstaff on a Building

A flagstaff stands on a building. From a point 20 m from the base of the building, the angles of elevation of the top of the building and the top of the flagstaff are 45° and 60° respectively. Find the height of the flagstaff.

Step 1: Let building = h, flagstaff = x. Point is 20 m away. For building top: tan 45° = h/20 → h = 20 m. For flagstaff top: tan 60° = (h+x)/20 → √3 = (20+x)/20 → 20+x = 20√3 → x = 20√3 − 20 = 20(√3−1) ≈ 14.64 m.

'Flagstaff height = height of (building + flagstaff) − height of building.'

Solving Strategy — Detailed

Draw a clear diagram: Mark all given angles and distances. Label the unknown as a variable.
Identify the right triangles: The vertical object (tower/building/tree) is ALWAYS perpendicular to the ground.
Select the correct ratio: tan θ = opposite/adjacent is most common. Use sin or cos only when hypotenuse appears.
Write equations: Set up one equation per observation. For multiple observations, you need multiple equations.
Solve systematically: Substitute and solve. Keep answers in exact form (with √, π) unless decimals are required.
Check reasonableness: Height should be positive. Does the answer make physical sense?

Common Mistakes — Extended

Confusing elevation and depression: Elevation is measured UP from horizontal. Depression is DOWN from horizontal. They are equal ONLY when the observer and target positions are swapped.
Forgetting observer height: When 'a person of height h m observes,' you MUST add h to the calculated height.
Unit mismatch: Ensure all distances use the SAME unit (usually metres) before applying trigonometric ratios.
Wrong trigonometric ratio: Height-distance problems almost always use tan. Use sin or cos only when the line of sight (hypotenuse) is given or required.
Incorrect diagram: The right angle is ALWAYS at the base of the vertical object.
Not rationalising denominators: Express final answers as (50√3)/3 rather than 50/√3.

AP Exam Focus

Topic	Marks	Frequency
Single observation (height or distance)	4	Very Common
Two observations (changing angle)	5	Common
Angle of depression problems	4	Common
Flagstaff on building	5	Common
Observer height included	4	Moderate

Self-Test Questions

A pole casts a 20 m shadow when the sun's elevation is 60°. Find the height of the pole. (Answer: 20√3 m)
From a cliff 150 m high, the angle of depression of a boat is 30°. Find the distance of the boat. (Answer: 150√3 m)
A ladder 10 m long reaches a window 8 m above ground. Find the angle of the ladder with the ground. (Answer: sin θ = 0.8, θ ≈ 53.13°)
Angle of elevation of a tower from a point 40 m away is 45°. Find the tower height. (Answer: 40 m)
From a building 75 m high, angles of depression of two objects are 30° and 60°. Find the distance between them (same side). (Answer: 50√3 m)
A person 1.8 m tall observes a tree top at 30° elevation from 25 m away. Find tree height. (Answer: 25/√3 + 1.8 ≈ 16.23 m)
A kite flies at 60 m height. The string makes a 60° angle with the ground. Find string length. (Answer: 60/sin 60° = 120/√3 = 40√3 ≈ 69.28 m)

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Heights and Distances

ANGLE OF ELEVATION: From a point on the ground, the angle between the horizontal and the LINE OF SIGHT to an object ABOVE. ANGLE OF DEPRESSION: From a height, the angle between the horizontal and the LINE OF SIGHT to an object BELOW. KEY RELATION: Angle of elevation from A to B = Angle of depression from B to A (alternate interior angles with the horizontal). FORMULAS: In a right triangle, tan θ = Opposite/Adjacent = height/horizontal distance. For SINGLE TRIANGLE: height h = d × tan θ (d = horizontal distance). FOR TWO-ANGLE PROBLEMS: Set up two equations with tan(angle₁) and tan(angle₂), two unknowns (height h, distance d or two distances d₁, d₂).

DIAGRAM DRAWING IS ESSENTIAL: (1) Draw a horizontal ground line. (2) Mark the vertical height (tower, building, cliff). (3) Mark the observer's position. (4) Draw the line of sight. (5) Mark the angle (elevation = from horizontal UP, depression = from horizontal DOWN). (6) Identify the right triangle. AP Board awards partial marks for correct diagrams even if calculation is wrong.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Using sin or cos where tan should be used

✓ In heights-and-distances problems, the HORIZONTAL DISTANCE and VERTICAL HEIGHT are the two legs of the right triangle — these are OPPOSITE and ADJACENT to the angle (not the hypotenuse). Therefore tan θ = height/distance is almost always the correct ratio. Sin and cos involve the hypotenuse (the slant distance/line of sight) — only use them if the line of sight length (hypotenuse) is given or required.

WATCH OUT

✗ Not equating the angle of depression with angle of elevation in two-triangle problems

✓ When an observer at the top of a cliff looks down at a boat: angle of depression from cliff = angle of elevation from boat to cliff top. These angles are EQUAL (alternate interior angles, since the observer's horizontal line and the ground are parallel). Using this equality lets you put the angle directly into the right triangle at ground level without any extra calculation.

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Applications of Trigonometry — Heights and Distances?

1 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

1 questions~2 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•ANGLE OF ELEVATION: From a point on the ground looking UP at an object above the horizontal. The angle is measured UPWARD from the horizontal line of sight. Example: standing 50 m from a tower, looking at the top — the angle between horizontal ground and the line of sight to the top.
•ANGLE OF DEPRESSION: From an elevated point looking DOWN at an object below the horizontal. The angle is measured DOWNWARD from the horizontal. Example: from a cliff top, looking at a boat in the sea — the angle between the horizontal and the line of sight to the boat.
•KEY EQUALITY: Angle of elevation from A to B = Angle of depression from B to A. These are ALTERNATE INTERIOR ANGLES between the horizontal line at A and the horizontal at B (which are parallel), cut by the line of sight. This lets you transfer the angle from one triangle to the other.
•STANDARD FORMULA: In a right triangle, tan θ = opposite/adjacent = height/horizontal distance. So HEIGHT = horizontal distance × tan θ. OR horizontal distance = height / tan θ. The hypotenuse (line of sight length) is found only if needed: line of sight = height / sin θ = distance / cos θ.
•SINGLE TRIANGLE PROBLEM (2 marks): One observer, one object. Draw one right triangle. Use tan θ = height/distance. One equation, one unknown. Standard setup: tower of height h, observer at distance d, elevation angle θ. tan θ = h/d.
•TWO-TRIANGLE PROBLEM (4 marks): Two observations from different positions OR two objects at different heights. Creates TWO right triangles sharing a common height or base. Write two equations with tan for the two angles. Solve the simultaneous equations. Most common setup: observer at two positions measuring angles to a tower top, OR observer seeing top and bottom of antenna on a building.
•WHEN TO USE SIN AND COS: In most AP SSC heights-and-distances problems, the GIVEN information is the horizontal distance (base) and you need the height — so tan is correct. Sin/cos are needed only when the LINE OF SIGHT length (hypotenuse, like a ladder leaning against a wall, or a cable wire) is given or required. Check: which two sides are given/needed? O and A → tan. O and H → sin. A and H → cos.
•STANDARD VALUES TO REMEMBER: tan 30° = 1/√3 ≈ 0.577. tan 45° = 1. tan 60° = √3 ≈ 1.732. For AP SSC: answers often come out to h = d/√3 (for 30°), h = d (for 45°), h = d√3 (for 60°). These clean answers suggest the setup is correct.
•COMPLEMENTARY ANGLES IN HEIGHTS: If angles from two positions are complementary (add to 90°), use the property tan α × tan(90°−α) = tan α × cot α = 1. From the two equations: h = d₁ tan α and h = d₂ tan(90°−α) = d₂ cot α = d₂/tan α → h × h = d₁ × d₂ → h = √(d₁d₂). The geometric mean of two distances gives the height.
•DIAGRAM CHECKLIST: (1) Horizontal ground line. (2) Vertical tower/building (perpendicular to ground). (3) Observer position(s). (4) Line(s) of sight. (5) Right angle marked at base of tower. (6) Angle(s) marked at observer position(s). Never start solving without this diagram — it converts the word problem into a geometry problem.

Andhra Pradesh (BIEAP) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Surveyors measuring building heights and land elevation

Professional surveyors use a theodolite (an instrument that measures horizontal and vertical angles precisely) to determine heights and distances. The process is exactly the AP SSC method: measure the horizontal distance to the base, measure the elevation angle to the top, compute height = distance × tan(elevation). Modern total stations (electronic theodolites) do this calculation automatically, but the underlying mathematics is identical to the Class 10 formula. Every building surveyed before construction uses this method.

Fire spotting towers and forest management

Forest fire watchers in towers observe fires at angles of depression and use the formula to estimate the distance to the fire: distance = tower height / tan(depression angle). Knowing the distance helps dispatch fire trucks on the correct route. AP's Forest Department uses this method in protected areas like the Nallamala Hills. The same principle applies to air traffic control towers estimating distance to approaching aircraft.

Cricket and sports analytics — trajectory angles

In cricket broadcasting, the angle of elevation of a ball's trajectory off the bat is measured using high-speed cameras. Combined with ball speed (from Hawk-Eye tracking), the height reached and landing distance are calculated using exactly these trigonometric relationships. A ball hit at 45° angle reaches maximum horizontal distance (when air resistance is neglected). Understanding heights and distances gives the mathematical foundation for understanding sports ball trajectories, long-jump distances, and shot-put angles.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Diagram FIRST — always draw before writing a single equation. The diagram earns partial marks even if the algebra is wrong. A clearly labelled diagram with right angles marked, angle values written, and sides labelled (h for height, d for distance) is worth 1–2 marks in AP SSC.
Label the right angle explicitly: mark the base of the tower/building/cliff as a right angle (90° or a small square). AP Board examiners check that the right triangle is correctly identified.
Two-triangle setup (4 marks): write tan(θ₁) = h/d₁ as equation (1) and tan(θ₂) = h/d₂ as equation (2). Label them explicitly. Then eliminate h or d by dividing or substituting. Show all algebraic steps — marks are awarded for each equation and each substitution.
Rationalise denominators: if your answer has √3 in denominator (e.g., 100/√3), write it as 100√3/3. AP Board mark schemes often require rationalised forms. Final answer: always write units (metres, cm, etc.).
Self-check with tan values: after getting h and d, verify that h/d equals the tan of the given angle. tan 30° = h/d should give 1/√3. If it gives a different value, there is an arithmetic error — recalculate.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Research triangulation — the surveying technique of measuring distances and areas by forming a network of triangles. Modern GPS and satellite positioning use triangulation extended to 3D (using four satellites to triangulate 3D position including altitude). Historically, the Great Trigonometrical Survey of India (1802–1871) measured the entire Indian subcontinent using triangulation — it discovered that Everest was the world's tallest mountain. Research how the GTS measured Everest's height from 160 km away using angles of elevation.
Investigate the tangent to a sphere — when a ship at sea sights the top of a lighthouse, the geometry is not a simple right triangle. The Earth curves, so the base of the triangle is actually a curve on the sphere's surface. Compute how far a lighthouse of height h can be seen on a spherical Earth of radius R: the sighting distance = √(2Rh) (using the Pythagorean theorem tangent to a sphere). At what height would a lighthouse need to be to be visible from 50 km away?
Research the Eratosthenes measurement of Earth's circumference (240 BCE) — Eratosthenes measured the shadow angle at noon in Alexandria when the sun was directly overhead in Syene (Aswan). The angle difference (7.2°, or 1/50 of a full circle) times the known ground distance (5000 stadia ≈ 800 km) gave Earth's circumference: 800 × 50 = 40,000 km (remarkably accurate — actual circumference: 40,075 km). This used the angle-of-elevation principle. Research how he did it and why it worked.
Explore parallax and astronomical distance measurement — astronomers measure the distance to nearby stars using PARALLAX: the apparent shift in a star's position when observed from opposite sides of Earth's orbit (6 months apart). The star appears to shift by an angle 2p (the parallax angle). Distance = 1/p parsecs (where p is in arcseconds). This is the cosmic equivalent of the two-observation heights problem in Class 10, scaled up by a factor of 10¹³. The first stellar parallax was measured by Friedrich Bessel in 1838.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

AP Board SSC (Class 10)Very High — Heights and Distances is a guaranteed 4–8 mark question in AP SSC Mathematics; one of the most predictable question types

JEE Main (Mathematics)Medium — Height and distance problems appear in JEE but are usually combined with more complex trigonometry (beyond standard angles)

NTSE (Mathematics)High — Heights and distances problems are standard NTSE Stage I and II questions

NDA / CDS (Mathematics section)High — Trigonometry applications including heights and distances are standard in NDA and CDS mathematics papers for defence services

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

ONE triangle: the problem gives ONE observation point and asks for height or distance. The single observer, single object setup. Setup: angle given, one unknown (height or distance) → one equation with tan → solve directly. TWO triangles: (1) Observer at TWO DIFFERENT POINTS (e.g., 'from point A, elevation is 30°; from point B, elevation is 60°') → Two triangles sharing the tower height. (2) Observer sees TWO objects at DIFFERENT heights (e.g., 'angles of depression to top and bottom of a lamp post') → Two triangles from the same elevated point. Any time you have two angle measurements, set up two equations and solve as simultaneous equations.

DIAGRAM: Vertical tower AB (A at top, B at base). Point P on ground. Angle at P = 30°. Right angle at B. tan(∠APB) = AB/PB → tan 30° = 100/PB → 1/√3 = 100/PB → PB = 100√3 ≈ 173.2 m. The point P is 100√3 m from the base. NOTE: tan 30° = 1/√3, NOT 1/3 — this is the most common computation error. Always write the exact value before computing.

DIAGRAM: Cliff AB = 30 m (A = top, B = base). Boat C on water level. Angle of depression from A = 60°. Since angle of depression from A = angle of elevation from C to A = 60° (alternate angles). In right triangle ABC: tan 60° = AB/BC = 30/BC → √3 = 30/BC → BC = 30/√3 = 30√3/3 = 10√3 ≈ 17.3 m. Note: we rationalise the denominator: 30/√3 = 30√3/3 = 10√3 — always rationalise unless a decimal is requested.

Complementary angles sum to 90°. In heights problems: if the elevation angles from two points to a tower are complementary (α and 90°−α), then tan(90°−α) = cot α = 1/tan α. Set up two equations: h = d₁ tan α (from nearer point) and h = d₂ × cot α = d₂/tan α (from farther point). Multiply both equations: h × h = d₁ tan α × d₂/tan α = d₁ × d₂ → h² = d₁ × d₂ → h = √(d₁d₂). This elegant result means you can find the height as the geometric mean of the two distances WITHOUT knowing the angle α. AP SSC tests this specifically.

The angle of depression θ satisfies tan θ = vertical height / horizontal distance (just like angle of elevation — the triangles are the same, only the observer's position is different). If height = 20 m and horizontal distance = 20 m, then tan θ = 20/20 = 1 → θ = 45°. If height = 30 m and distance = 30/√3 = 10√3 m, then tan θ = 30/(10√3) = 3/√3 = √3 → θ = 60°. Recognise these: tan 30° = 1/√3, tan 45° = 1, tan 60° = √3. If tan θ = 1/√3, then θ = 30°.

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