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Class 10 Mathematics Chapter 9 of 15

Chapter 9 — Some Applications Of Trigonometry

Overview

Chapter: Some Applications of Trigonometry (NCERT Class 10) — Introduction, importance, key themes, and learning outcomes. This chapter develops the practical side of right-triangle trigonometry by showing how sine, cosine and tangent are used to find heights and distances in real-life situations. Importance: it links abstract trigonometric ratios to concrete problems encountered in surveying, navigation, architecture, engineering and everyday life (e.g., heights of towers, angles of elevation/depression, ladder problems and shadows). Key themes: angle of elevation and depression; modelling a physical situation by a right triangle; using trigonometric ratios (sin, cos, tan) and complementary-angle relations; application of the Pythagorean theorem where required; strategies for multi-step problems (two observations, successive points). What the student will learn: how to translate word problems into labelled right triangles, choose the correct trig ratio, compute unknown lengths or heights using given angles (or vice versa), apply complementary-angle relations to simplify problems, solve standard NCERT problem types such as single-observer height, two-observer distance/height,…

Learning Objectives

  • Define the trigonometric ratios sine, cosine and tangent for acute angles in a right-angled triangle
  • Explain the relationships between trigonometric ratios of complementary angles (co-function identities)
  • Use exact values of trigonometric ratios for 30°, 45° and 60° in problem solving
  • Apply the identity tanθ = sinθ/cosθ and Pythagorean relations to relate sides and ratios
  • Solve problems to determine heights and distances using angles of elevation and depression
  • Apply inverse trigonometric functions to find angles from given trigonometric ratios
  • Model real-life situations (trees, towers, poles, shadows, ladders, ships) by constructing right triangles
  • Formulate and solve multi-step problems involving two or more right triangles

Topics in this chapter

10 topics · tap a topic title to jump straight to it.

🔢1

Introduction

Some Applications of Trigonometry introduces how basic trigonometric ratios (sin, cos, tan) are used to find unknown heights and distances that cannot be measured directly. The chapter focuses on right-triangle modelling of real situations, definition and use of angle of elevation and angle of depression, and standard problem-solving steps.

Key ideas:

  • Angle of elevation: the angle formed by the line of sight when an observer looks upward from a horizontal line to an object.
  • Angle of depression: the angle formed by the line of sight when an observer looks downward from a horizontal line to an object. (Angle of depression from the top = angle of elevation from the bottom.)
  • Model each situation by a right triangle: vertical object = height (opposite), horizontal distance from object = base (adjacent), angle measured from horizontal = θ. Then apply trigonometric ratios.
  • Typical procedure: draw a clear diagram, mark the angle and known lengths, identify the appropriate trig ratio, write the equation, and solve for the unknown.

Assumptions usually made in problems: the ground is horizontal, the object (tower, tree) is vertically straight, and angles are measured with respect to the horizontal.

Common derived result used often: when the same top point is observed from two points on the ground separated by a known distance, you can use the two angles of elevation to find the height without measuring directly (see formula in the formulas list).

📌 Examples
  • Example 1: Height of a tower. From a point 40 m from the foot of a tower the angle of elevation to the top is 30°. Height h = 40 × tan 30° = 40 × (1/√3) ≈ 23.09 m.
  • Example 2: Width of a river. From a point A on one bank, angle of elevation to a tree on the opposite bank is 45°. From point B, 20 m further back from A, angle is 30°. Let width be w. Using two right triangles with same height, w can be found by applying tan relations (or using the two-position formula).
  • Example 3: Angle of depression. From the top of a lighthouse 50 m high, the angle of depression to a boat is 20°. Horizontal distance d = 50 / tan 20° ≈ 137.4 m.
  • Example 4: Two-position measurement for height. An observer measures angles of elevation 35° and 55° at two points on the same line toward a tower; distance between the two points is 30 m (closer point gives 55°). Height h = 30 × tan35° × tan55° / (tan55° − tan35°) (compute numerically).
🧮 Formulas
  1. Definitions in a right triangle (θ angle at base): sin θ = opposite / hypotenuse, cos θ = adjacent / hypotenuse, tan θ = opposite / adjacent
  2. tan θ = sin θ / cos θ
  3. Angle of elevation = Angle of depression (when measured between horizontal and line of sight to the same object from two opposite points).
  4. Height from single observation: if angle of elevation = θ and horizontal distance = d, then height h = d × tan θ
  5. Two-position (two angles θ1, θ2 from two points separated by distance D along the same line toward the object; θ2 > θ1, nearer point has θ2): h = D × tan θ1 × tan θ2 / (tan θ2 − tan θ1)
  6. Rearranging for distance: if h and θ known, horizontal distance d = h / tan θ
📊 Visual ideas
Sketch 1 (single-observation diagram): draw a right triangle with vertical side labelled h (height), horizontal base labelled d (distance from observer to foot), and angle θ at the observer between base and hypotenuse (line of sight). Label tan θ = h/d and show formula h = d tan θ.
Sketch 2 (two-position diagram): draw a vertical line representing the object of height h. Mark two points A and B on the same horizontal line at distances x and x + D from the foot respectively, with angles of elevation θ2 (at nearer point) and θ1 (at farther point). Show two right triangles sharing the same vertical side and derive h = D tan θ1 tan θ2 / (tan θ2 − tan θ1).
Sketch 3 (angle of depression): draw the top of a tower, a horizontal line from the top, and a line of sight downwards making angle of depression θ to a point on ground. Show that angle of depression equals angle of elevation from ground point and form right triangle to compute d or h.
Sketch 4 (trig-function behavior): plot tan θ vs θ for 0° < θ < 90° to visualize how tan θ increases from 0 to +∞ — useful to anticipate solution behavior when angles approach 90°.
📐2

Review of Right‑Triangle Trigonometry

What it is: Right‑triangle trigonometry studies the relationships between the angles and side lengths of a triangle that has one 90° angle. For an acute angle θ in a right triangle, the three primary trigonometric ratios—sine, cosine and tangent—are defined as ratios of two sides.

Basic definitions (for angle θ):

  • Opposite — side opposite angle θ.
  • Adjacent — side next to θ (not the hypotenuse).
  • Hypotenuse — the side opposite the right angle (longest side).

Primary ratios:

  • sin θ = (opposite)/(hypotenuse)
  • cos θ = (adjacent)/(hypotenuse)
  • tan θ = (opposite)/(adjacent) = sin θ / cos θ

Reciprocal ratios: cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ.

Key identities: In any right triangle with angle θ,

  • sin^2 θ + cos^2 θ = 1 (Pythagorean identity)
  • 1 + tan^2 θ = sec^2 θ
  • tan θ = sin θ / cos θ
  • sin(90° − θ) = cos θ, cos(90° − θ) = sin θ (complementary angles)

How to use them to solve problems: Given one acute angle and one side, you can find other sides using the appropriate ratio. Given two sides, you can find an acute angle using inverse trig functions (e.g., θ = sin^−1(opposite/hypotenuse)). For many real problems (heights and distances) you model the situation as a right triangle and apply these ratios.

Angle of elevation and depression: If an observer looks up to a point, the angle between the horizontal and the line of sight is the angle of elevation; if looking down, it is the angle of depression. The same trig ratios apply because the formed triangles are right triangles.

Practical tips: Always identify the angle of interest, label opposite/adjacent/hypotenuse, choose the correct ratio, and solve algebraically. Check units (degrees/radians) in calculators: CBSE problems use degrees.

📌 Examples
  • Height of a tower: From a point 30 m from the foot of a tower the angle of elevation of the top is 60°. Find the height. (Use tan 60° = √3 → height = 30 × √3 ≈ 51.96 m.)
  • Ladder problem: A ladder 13 m long leans against a wall making an angle of 60° with the ground. How high up the wall does it reach? (Use sin 60° = h/13 → h = 13 × (√3/2) ≈ 11.26 m.)
  • Width of a river: From two points A and B on one bank, 100 m apart, the angles of elevation to a tree on the opposite bank are 30° and 60°. Using two right triangles and trig ratios, compute the distance from B to the tree (set up equations using tan 30° and tan 60° and subtract).
  • Shadow problem: At a certain time the shadow of a pole is 10 m long and the angle of elevation of the sun is 30°. Find the pole's height. (tan 30° = height/10 → height = 10 × tan 30° = 10 × (1/√3) ≈ 5.77 m.)
  • Finding an angle: In a right triangle, if opposite = 7 and hypotenuse = 25, then θ = sin^−1(7/25) ≈ 16.26°.
🧮 Formulas
  1. sin θ = opposite / hypotenuse
  2. cos θ = adjacent / hypotenuse
  3. tan θ = opposite / adjacent = sin θ / cos θ
  4. cosec θ = 1 / sin θ, sec θ = 1 / cos θ, cot θ = 1 / tan θ
  5. sin^2 θ + cos^2 θ = 1
  6. 1 + tan^2 θ = sec^2 θ
📊 Visual ideas
Plot y = sin θ and y = cos θ for θ from 0° to 90°. Show that sin θ starts at 0 and increases to 1, while cos θ starts at 1 and decreases to 0. Mark points at 30°, 45°, 60° with their exact values.
Plot y = tan θ for θ from 0° to 80°. Show the increasing curve and note the steep rise as θ approaches 90° (vertical asymptote).
Draw a labeled right triangle and overlay the definitions: mark opposite, adjacent, hypotenuse and show sin, cos, tan as ratios (use a slider to change θ and observe side-length ratios change).
Bar chart comparing sin, cos, tan values at 30°, 45° and 60° to help students memorise key values.
📐3

Angle of Elevation and Angle of Depression

Definitions

An angle of elevation is the angle made by the line of sight with the horizontal when an observer looks upward at an object. An angle of depression is the angle made by the line of sight with the horizontal when an observer looks downward at an object.

Key idea

  • Draw the horizontal through the observer's eye. The angle between this horizontal and the line joining the observer's eye to the object is the required angle.
  • In typical problems you form a right triangle: one leg is the horizontal distance between observer and object, the other is the vertical difference in heights.

Why angle of elevation = angle of depression

If an observer at point O looks up to top T of an object and looks down to point B on the ground, draw the horizontal line through O. The angle between OT and the horizontal is the angle of elevation and the angle between OB and that same horizontal (on the other side) is the angle of depression. By alternate interior angles (parallel horizontal lines), the angle of elevation measured from the ground level point is equal to the angle of depression measured from the observer — in practice this means: when you connect the object and observer you get congruent right triangles that share the same acute angle.

How to solve problems

  • Identify observer height and object height (or their difference). Mark the horizontal and form the right triangle.
  • Use trig ratios: sinθ = opposite/hypotenuse, cosθ = adjacent/hypotenuse, tanθ = opposite/adjacent. For height-distance problems tan is most commonly used: tanθ = (vertical difference)/(horizontal distance).
  • Solve for the unknown (height, distance or angle). If solving for an angle use θ = arctan(opposite/adjacent).

Special notes

  • If the observer's eye is not at ground level, incorporate the observer's height: tree height = observer height + (horizontal distance)·tan(angle of elevation).
  • When two objects of different heights are involved, use two right triangles and their common horizontal distance; subtract or add heights appropriately.
📌 Examples
  • Example 1 — Height of a tower: An observer 50 m from the foot of a tower measures an angle of elevation 30°. Find the height of the tower. Solution: tan30° = height / 50. So height = 50·tan30° = 50·(1/√3) ≈ 28.87 m.
  • Example 2 — Distance to a boat: From the top of a cliff 50 m high, the angle of depression to a boat is 30°. How far is the boat from the base of the cliff (horizontal distance)? Solution: angle of depression = 30° ⇒ angle of elevation from boat = 30°. tan30° = 50 / distance ⇒ distance = 50 / tan30° = 50·√3 ≈ 86.60 m.
  • Example 3 — Include observer height: A person 1.6 m tall stands 10 m from a tree and measures the angle of elevation to the top of the tree as 45°. Find the tree's height. Solution: tan45° = (tree_height − 1.6) / 10 ⇒ 1 = (tree_height − 1.6)/10 ⇒ tree_height = 10 + 1.6 = 11.6 m.
  • Example 4 — Two towers: Two towers stand 80 m apart. From the top of the taller tower (height 50 m), the angle of depression to the top of the shorter tower is 15°. Find the height of the shorter tower. Solution: Let shorter height = h. Vertical drop = 50 − h. tan15° = (50 − h)/80 ⇒ 50 − h = 80·tan15° ⇒ h = 50 − 80·tan15° ≈ 50 − 21.42 ≈ 28.58 m.
🧮 Formulas
  1. sin θ = opposite / hypotenuse
  2. cos θ = adjacent / hypotenuse
  3. tan θ = opposite / adjacent
  4. For height-distance problems: tan(θ) = (vertical difference in heights) / (horizontal distance)
  5. If observer's eye is at height h_o and horizontal distance is d and angle of elevation is θ: object height = h_o + d·tan θ
  6. Angle of elevation from one point equals angle of depression from the other when measured to the same line of sight (use alternate interior angles)
📊 Visual ideas
Simple right-triangle diagram for angle of elevation: draw ground horizontal, mark observer O at ground level, mark object top T vertically above foot A. Label distance OA = d, height AT = h. Draw line OT and angle ∠O (between OT and horizontal) = angle of elevation θ. Annotate tanθ = h/d.
Angle of depression from a cliff: draw cliff top point C (observer eye), horizontal line through C, ground point B at base of cliff, boat at point P on water. Angle between horizontal at C and line CP is angle of depression α. Draw right triangle CPB, label height CB = H, horizontal BP = x. Use tanα = H/x. Note equality of α with angle of elevation from P to C.
Two-level observer (observer height ≠ 0): draw observer eye at height h_o above ground at O, tree top T at distance d from base. Triangle with vertical (T − ground) minus h_o as opposite side. Label and use tree_height = h_o + d·tan(θ).
Coordinate-style sketch for classroom plotting: place observer at origin (0,h_o), horizontal axis x, ground as x-axis. Draw line of sight making angle θ above horizontal and intersecting vertical line x = d at height h. Compute h = h_o + d·tan θ. This helps plot multiple sightlines (vary θ) to see how height changes with angle.
🔢4

Heights and Distances — Single Observation Problems

What are single observation problems? These are trigonometry problems where you make one measurement of an angle (angle of elevation or depression) from a single point and use it to find an unknown height or horizontal distance. They are solved by modelling the situation as a right triangle and using the trigonometric ratios.

Key ideas

  • Angle of elevation — the angle between the horizontal line of sight and the line of sight upward to the top of the object.
  • Angle of depression — the angle between the horizontal line of sight and the line of sight downward to the object. (Angle of depression from a point equals the angle of elevation of that point from the object.)
  • Assumptions — ground between observer and object is level and distances are horizontal. If the observer’s eye is at a nonzero height above the ground, include that height in the final calculation.

Method / Steps to solve

  1. Draw a clear diagram showing the observer, the top and base of the object, and a horizontal line through the observer.
  2. Identify the right triangle: the vertical side is the object’s height above the observer’s horizontal line (or above ground), the horizontal side is the horizontal distance between observer and object base.
  3. Write the trigonometric ratio that involves the given angle and the two sides (commonly tan θ = opposite / adjacent).
  4. Solve algebraically for the unknown. If the observer’s eye is at height h_e above ground, and the calculated vertical from observer to top is h_rel, then object height H = h_e + h_rel (or subtract if measuring below).
  5. Check units and round off sensibly.

Common formulas used (from a single observation)

  • If angle of elevation = θ, horizontal distance = d, and observer’s eye is at ground level: height of object H = d · tan θ.
  • If observer’s eye is at height h_e above ground and angle of elevation to the top is θ: H = h_e + d · tan θ.
  • If angle of depression = θ from top of known height H (or from eye height h_e): horizontal distance d = (H − h_e) / tan θ (or H / tan θ when measuring from the top of the object).

Tips / common mistakes

  • Always use the horizontal distance (adjacent side) — not the slant distance — in tan calculations.
  • Keep units consistent (metres with metres, etc.).
  • Pay attention to whether the angle was measured from the observer’s eye level or from ground level; add or subtract the eye height as needed.
📌 Examples
  • Example 1 — Tree height (observer at ground level): From a point 20 m from a tree, the angle of elevation to the top is 30°. Find the tree height. Solution: H = d·tanθ = 20·tan30° = 20·(1/√3) ≈ 11.55 m.
  • Example 2 — Building height (observer eye height not zero): An observer whose eye is 1.6 m above ground measures an angle of elevation 45° to the top of a building. If the horizontal distance to the building is 10 m, find the building height. Solution: additional height above eye = d·tanθ = 10·tan45° = 10·1 = 10 m. So total H = 1.6 + 10 = 11.6 m.
  • Example 3 — Angle of depression: From the top of a lighthouse 50 m high, the angle of depression to a boat is 30°. Find the horizontal distance of the boat from the lighthouse. Solution: tan30° = opposite/adjacent = 50/d ⇒ d = 50 / tan30° = 50·√3 ≈ 86.6 m.
🧮 Formulas
  1. tan θ = (opposite side) / (adjacent side)
  2. If angle of elevation θ and horizontal distance d (observer at ground): H = d · tan θ
  3. If observer eye height h_e and angle of elevation θ: H = h_e + d · tan θ
  4. If angle of depression θ from a point at height H: d = H / tan θ (or d = (H − h_e) / tan θ if eye height h_e is given)
📊 Visual ideas
Right-triangle diagram: horizontal axis (ground), vertical axis (object height). Mark observer point, base of object, top of object, angle θ at observer, label adjacent = d and opposite = h (or h_rel). This visual is essential for every single-observation problem.
Graph of H = d·tanθ for fixed d: plot θ (0° to, say, 80°) on x-axis and H on y-axis. This shows H increases slowly for small θ then rapidly as θ approaches 90° (vertical asymptote). Useful to illustrate sensitivity of height to angle near 90°.
Graph of d = H / tanθ for fixed H: plot θ on x-axis and d on y-axis. This is a decreasing curve; small angles produce large horizontal distances. Helpful to show why small measurement errors in small angles give large errors in distance.
🔢5

Heights and Distances — Two Observation Problems

Concepts: Heights and distances problems use right-triangle trigonometry to find the height of an object (tower, tree, building) or horizontal distances when one or more angles of elevation or depression are known.

Definitions:

  • Angle of elevation: angle between the horizontal line of sight and the line of sight to an object above the horizontal.
  • Angle of depression: angle between the horizontal line of sight and the line of sight to an object below the horizontal (equal to the corresponding angle of elevation from the lower point).
  • Basic relation: tan(θ) = (height)/(horizontal distance).

General approach for two-observation problems (two observations of the top of the same object from two different points):

  1. Draw the vertical height h and the two horizontal distances from the foot of the object to the observation points (d1, d2). Mark the measured angles of elevation α and β at the observation points.
  2. Write tan relations: tan α = h / d1, tan β = h / d2.
  3. If the distance L between the two observation points is known, express L in terms of d1 and d2 (either d2 − d1 or d1 + d2 depending on whether the points lie on the same side or opposite sides of the object), then eliminate d1, d2 to solve for h.

Two common cases and derivations:

Case A — observation points on the same side of the object (d2 > d1). Let the nearer point see angle α (larger) and the farther see β (smaller). The distance between points is L = d2 − d1. From tan relations:

h = d1·tan α = d2·tan β, hence d1 = h / tan α, d2 = h / tan β.

So L = h(1/tan β − 1/tan α) = h·(tan α − tan β)/(tan α·tan β).

Solving for h gives the formula:

h = L · (tan α · tan β) / (tan α − tan β)

Case B — observation points on opposite sides of the object. Distances from foot add: L = d1 + d2. Using d1 = h / tan α and d2 = h / tan β:

L = h(1/tan α + 1/tan β) = h·(tan α + tan β)/(tan α·tan β).

So:

h = L · (tan α · tan β) / (tan α + tan β)

Note: Angle of depression problems are handled the same way because the angle of depression from the top equals the angle of elevation from the lower point.

Practical tips: always check which angle is larger (the nearer point gives larger angle), choose α and β consistently, ensure angles are in the same units (degrees) and use the correct sign in the denominator (minus for same-side, plus for opposite-side).

📌 Examples
  • Example 1 (same side): From two points on the same straight line towards a tower the angles of elevation of the top are 60° and 30°. The distance between the two observation points is 40 m. Find the height h of the tower. Solution: tan60 = √3 ≈ 1.732, tan30 ≈ 0.5774. Use h = L · (tanα·tanβ)/(tanα − tanβ) = 40 · (1.732·0.5774)/(1.732 − 0.5774) ≈ 40 · 1 / 1.1546 ≈ 34.6 m.
  • Example 2 (opposite sides): Two observers stand on opposite banks of a river (width L = 50 m). They measure angles of elevation to the top of a vertical mast as 45° and 30°. Find the mast height. Solution: tan45 = 1, tan30 ≈ 0.5774. Use h = L · (tanα·tanβ)/(tanα + tanβ) = 50 · (1·0.5774)/(1 + 0.5774) ≈ 50 · 0.3663 ≈ 18.3 m.
  • Example 3 (angle of depression): From the top of a lighthouse, angles of depression to two buoys on the same side are 20° and 10°. The buoys are 200 m apart. Find the height of the lighthouse. Treat angles of depression as angles of elevation from the buoys and use the same-side formula with α = 20°, β = 10°, L = 200 m, compute tan values and apply h = L·(tanα·tanβ)/(tanα − tanβ).
🧮 Formulas
  1. tan θ = (opposite) / (adjacent) = height / horizontal distance
  2. If two observations are on the same side (distance between points L = d2 − d1): h = L · (tan α · tan β) / (tan α − tan β)
  3. If two observations are on opposite sides (distance between points L = d1 + d2): h = L · (tan α · tan β) / (tan α + tan β)
  4. Angle of elevation = angle of depression (corresponding angles), so depression problems convert to elevation problems
📊 Visual ideas
Sketch 1: Vertical line for the object of height h at origin. Two horizontal points at distances d1 and d2 to the right (same side case). Draw right triangles from each point to the top, label angles α (nearer) and β (farther), label L = d2 − d1. Annotate tan relations and display the derived formula beside the sketch.
Sketch 2: Vertical object with one point to the left and one to the right (opposite sides case). Draw two right triangles meeting at the top, label angles α and β and L = d1 + d2. Show tan relations and formula for h.
Interactive graph idea: slider-controlled α and β with fixed L; dynamically plot computed h(α,β) and show the two triangles updating. Also include plots of tan(θ) curves and a small panel showing intermediate values (tan α, tan β, numerator and denominator) to visualize sensitivity.
Auxiliary plot: For fixed L, plot h vs α for a few fixed β values (or vice versa) to illustrate how h changes with the observation angles.
🔢6

Combined and Mixed Observation Problems

What the topic means
Combined and mixed observation problems involve finding the height (or distance) of an object by using two different observations (two angles of elevation/depression) taken from two different points. The two observation points are usually a known distance apart. By forming two right triangles and using trigonometric ratios (mainly tan), we get two equations which we solve simultaneously.

Basic setup
Let the vertical height to be found be h and the foot of the object be T on the ground. Two observers stand at A and B on the ground, with AB = d. Let the angles of elevation of the top from A and B be α and β respectively.

Two common configurations and their derivations

1) Observations from the same side of the object
Suppose A and B lie on the same side of T, with A nearer to T so that AT = x and BT = x + d. From right triangles:

h = x·tanα and h = (x + d)·tanβ.

Eliminate x: x(tanα − tanβ) = d·tanβ, so x = d·tanβ/(tanα − tanβ).

Thus h = x·tanα = d·tanα·tanβ/(tanα − tanβ).

2) Observations from opposite sides of the object
If A and B are on opposite sides of T, let AT = x and BT = y so that d = x + y. From right triangles:

h = x·tanα and h = y·tanβ ⇒ x = h/tanα, y = h/tanβ.

So d = h(1/tanα + 1/tanβ) = h(tanα + tanβ)/(tanα·tanβ), and hence

h = d·tanα·tanβ/(tanα + tanβ).

Mixed observations
Mixed problems are those where angles of elevation and depression (or observations with nonzero eye-height) are combined. Key points:

  • An angle of depression from a top equals the angle of elevation from the point on the ground.
  • If the observer's eye is at height s (not at ground level), replace h by (h − s) or (h + s) depending on geometry (usually use h − s when computing vertical difference to ground point observed).

Important cautions

  • If tanα = tanβ in the same-side formula the denominator is zero: that corresponds to equal angles (observers equidistant) and the simple formula fails (special handling required).
  • Angles must be expressed in the same units (degrees or radians) and tangents must be computed consistently.
📌 Examples
  • Example 1 (same side): Two points A and B are 20 m apart on the same side of a tower. Angles of elevation of the top from A and B are 60° and 30° respectively. Find the height h of the tower. Solution: h = d·tanα·tanβ/(tanα − tanβ). Here tan60 = √3, tan30 = 1/√3 so tan60·tan30 = 1 and tan60 − tan30 = (√3 − 1/√3) = 2/√3. Thus h = 20·1/(2/√3) = 20·(√3/2) = 10√3 ≈ 17.32 m.
  • Example 2 (opposite sides): Two observers on opposite sides of a vertical pole are 30 m apart. The angles of elevation to the top are 30° and 45°. Find the pole's height. Solution: h = d·tanα·tanβ/(tanα + tanβ). With tan30 ≈ 0.5774, tan45 = 1, h ≈ 30·0.5774/(1.5774) ≈ 10.98 ≈ 11.0 m.
  • Example 3 (mixed: angles of depression): From the top of a lighthouse two ships on opposite sides are observed with angles of depression 30° and 45°. The ships are 50 m apart. Find the height of the lighthouse. (Angles of depression equal angles of elevation from the ships.) Use opposite-sides formula: h = 50·tan30·tan45/(tan30 + tan45) ≈ 50·0.5774/(1.5774) ≈ 18.3 m.
🧮 Formulas
  1. Same side (A and B on same side, AB = d, angles α at nearer A and β at farther B): h = d·tanα·tanβ / (tanα − tanβ) (valid when tanα ≠ tanβ).
  2. Opposite sides (A and B on opposite sides, AB = d): h = d·tanα·tanβ / (tanα + tanβ).
  3. If observer eye-height = s (not at ground): replace h by (h − s) in triangle relations. For example, for opposite sides when observations are to the top from ground level but the observer is at height s, use (h − s) in place of h inside formulas.
  4. Angle of depression from top = corresponding angle of elevation from ground point (use this to convert depression problems to elevation problems).
📊 Visual ideas
Diagram 1 (Same side geometry): Draw a vertical line for the tower of height h at point T. Mark two ground points A (distance x from T) and B (distance x + d from T) on the same side. Draw right triangles ATA_top and BTB_top showing angles α at A and β at B. Label d = AB. This static diagram helps visualize derivation of h = d·tanα·tanβ/(tanα − tanβ).
Diagram 2 (Opposite sides geometry): Draw vertical line at T and ground points A and B on opposite sides at distances x and y (so AB = x + y = d). Draw right triangles showing angles α and β and label tan relations. Use arrows to show d = x + y and derive h = d·tanα·tanβ/(tanα + tanβ).
Interactive plot suggestion: For fixed d and fixed β, plot h(α) = d·tanα·tanβ/(tanα − tanβ) as α varies (same-side case). This shows h → ∞ as α → β (vertical asymptote) and helps students see sensitivity when angles are close. Add sliders for d and β.
Visualization for mixed problems: Sketch the observer at height s on top of a tower and two ground points with angles of depression. Convert the depression angles to elevations and use the opposite/same side diagrams. Optionally overlay numeric values and show step-by-step triangle heights.
🔢7

Practical Word Problems and Models

What this topic means

In this topic we convert real-life situations (towers, trees, ladders, shadows, ships seen from cliffs, etc.) into right‑triangle models and use trigonometric ratios (sin, cos, tan) to find unknown lengths or angles. The key idea is: draw a clear diagram, identify right triangles, label the known quantities, choose appropriate trigonometric ratio(s), write equations, and solve.

Common modelling steps

  1. Draw a neat diagram and mark the angle(s) of elevation or depression and known distances.
  2. Replace the real-life object by a right triangle (height, base, hypotenuse as needed).
  3. Use definitions: sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent.
  4. Form equations, solve for the unknown, check units and reasonableness.

Typical situations

  • Single-angle problems: given distance on ground and angle of elevation to top → find height: h = d · tan θ.
  • Angle of depression problems: angle of depression from top equals angle of elevation from ground; treat similarly.
  • Two-point problems: same object observed from two points on the same line (separated by known distance) giving two angles → derive h = d · tan A · tan B /(tan A − tan B) (when A > B), where d is the distance between the observation points.
  • Ladder problems: ladder length L leaning against a wall reaches height h = L · sin θ (θ = angle with horizontal), horizontal distance from wall = L · cos θ.
  • Shadow problems: if a vertical object of height h casts shadow length s under sun elevation θ, then tan θ = h/s → h = s · tan θ.

Notes and cautions

  • Always check which angle is with ground (angle of elevation) or with horizontal from top (angle of depression).
  • Ensure angles are measured in degrees (or radians consistently) and that tan A ≠ tan B when using two-point formula.
  • Sketching is essential — a correct label often prevents algebra mistakes.
📌 Examples
  • Example 1 — Height of a tower: From a point 50 m from the foot of a tower the angle of elevation is 30°. Find the tower's height. Solution: tan 30° = h/50 ⇒ h = 50 · tan 30° = 50 · (1/√3) ≈ 28.87 m.
  • Example 2 — Two observers: Two observers stand on the same straight line at distances d = 200 m apart from each other and observe the top of a tower. The nearer observer sees the top at 60° and the farther observer at 30°. Find the tower height. Solution: Use h = d·tan A·tan B/(tan A − tan B) with A = 60°, B = 30°. tan60 = √3 ≈ 1.732, tan30 ≈ 0.57735. Compute h = 200·(1.732·0.57735)/(1.732−0.57735) = 200·1/(1.1547) ≈ 173.21 m.
  • Example 3 — Shadow problem: A pole casts a shadow 5 m long. If the sun's elevation angle is such that tan θ = 2, find the pole's height. Solution: h = s · tan θ = 5 · 2 = 10 m. If you instead know h and s, you can find θ = arctan(h/s) ≈ arctan(2) ≈ 63.43°.
  • Example 4 — Ladder against wall: A ladder 13 m long leans against a wall and reaches a height of 12 m on the wall. Find the angle θ the ladder makes with the ground and the horizontal distance from the wall to the foot of the ladder. Solution: sin θ = opposite/hypotenuse = 12/13 ⇒ θ = arcsin(12/13) ≈ 67.38°. Horizontal distance = 13 · cos θ = √(13^2 −12^2) = 5 m.
  • Example 5 — Angle of depression: From the top of a lighthouse 60 m high, the angle of depression to a boat is 30°. Find the horizontal distance of the boat from the base of the lighthouse. Solution: angle of depression = 30° ⇒ tan 30° = 60/d ⇒ d = 60 / tan 30° = 60 · √3 ≈ 103.92 m.
🧮 Formulas
  1. sin θ = opposite / hypotenuse
  2. cos θ = adjacent / hypotenuse
  3. tan θ = opposite / adjacent
  4. Height from one observation: h = d · tan θ (d = horizontal distance from object)
  5. Ladder: height reached h = L · sin θ ; base distance = L · cos θ
  6. Shadow: tan θ = height / shadow_length ⇒ height = shadow_length · tan θ
📊 Visual ideas
Sketch 1: Single right triangle — horizontal base labeled d, vertical height h, angle θ at base. This visual shows tan θ = h/d. Good for tower/tree examples.
Sketch 2: Two-triangle model for two-observer problems — draw tower at one end, two observation points on base line separated by distance d; draw two right triangles sharing the same vertical height h and label angles A (nearer) and B (farther). Use to derive and visualize h = d·tanA·tanB/(tanA−tanB).
Sketch 3: Ladder triangle — hypotenuse L leaning to wall, vertical h and base x. Mark θ between ladder and ground; show h = L·sinθ and x = L·cosθ.
Sketch 4: Shadow & sun rays — vertical object height h, horizontal shadow s, ray from top making angle θ with horizontal. Use to show tan θ = h/s.
🔢8

Problem‑Solving Strategy and Diagramming

Overview
Problem‑Solving Strategy and Diagramming is the systematic way to convert a word problem into a clear geometric diagram (usually a right triangle or set of right triangles), choose appropriate trigonometric ratios, form equations, and solve for the unknowns. Careful diagramming reduces mistakes and makes use of basic identities and similar triangles straightforward.

Step‑by‑step strategy

  1. Read and understand: Identify what is given (angles, distances, heights) and what must be found.
  2. Visualize and draw: Sketch the situation roughly first. Then draw a neat, labelled diagram showing horizontal, verticals, given distances and angles.
  3. Introduce right angles: If the figure is not an explicit right triangle, draw a perpendicular (height from top to ground or horizontal from a point) to create right triangles.
  4. Label carefully: Mark angles (θ), sides (height h, distance d), and right angles. Mark known lengths and unknowns with symbols.
  5. Choose trigonometric ratio: For an acute angle θ in a right triangle pick sinθ, cosθ, or tanθ depending on which sides (opposite, adjacent, hypotenuse) are known/unknown.
  6. Write equations and solve: Use ratios (e.g. tanθ = opposite/adjacent), Pythagoras or similar triangles as needed; solve algebraically for the unknown; check units and reasonableness.
  7. Verify: Check the answer by plugging back into the diagram or by checking limiting cases (e.g. angles 0°, 90°) and units.

Diagramming tips

  • Always draw the horizontal line (eye level/ground) when dealing with elevation/depression problems.
  • Angle of elevation (from observer looking up) is measured above horizontal; angle of depression is measured below horizontal. Angle of depression from top equals angle of elevation from the ground point to the top (alternate interior angles) when lines are parallel.
  • When two observers or two sight lines are involved, draw both right triangles sharing a common vertical (height); this often yields two equations in the same unknown height.
  • Label unknown distances as variables (x, d1, d2) rather than immediately introducing numbers that could confuse algebra steps.
  • Mark right angles and show small arcs for angles you use in trigonometric ratios — this clarifies which side is opposite/adjacent.

Common mistakes to avoid

  • Mixing up opposite and adjacent with respect to the given angle.
  • Forgetting that angle of depression is measured from the horizontal (not from the vertical).
  • Not introducing a perpendicular when the object’s top and base do not align vertically in the rough sketch.
  • Ignoring units (meters vs kilometres) or failing to check the reasonableness of an answer (e.g., height larger than expected).

How trigonometry and similar triangles work together
Trigonometric ratios (sin, cos, tan) give relations between an angle and side lengths in a right triangle. Similar triangles often appear when two observers at different distances see the same top — the two right triangles are similar, so corresponding sides are proportional, which leads to equations involving tan of the two angles.

📌 Examples
  • Example 1 — Height of a tree: An observer stands 10 m from a tree and measures the angle of elevation to the top as 30°. Find the tree height. Solution: Draw right triangle with base 10 m and angle 30°. Use tan30° = height/10. So height = 10 * tan30° = 10 * (1/√3) ≈ 5.77 m.
  • Example 2 — Two observers, same object: Two observers at distances 20 m and 10 m from a tower measure angles of elevation 30° and 60° respectively. Find the height of the tower. Solution: For the nearer observer: h = 10 * tan60° = 10 * √3 ≈ 17.32 m. (Alternatively, both must give same h so either triangle yields same result.)
  • Example 3 — Angle of depression: From the top of a lighthouse 40 m high, the angle of depression to a boat is 25°. How far is the boat from the foot of the lighthouse? Solution: Angle of depression equals angle of elevation from the boat. So tan25° = 40/d ⇒ d = 40 / tan25° ≈ 40 / 0.4663 ≈ 85.8 m.
  • Example 4 — Ladder leaning against wall: A ladder of length 13 m leans against a vertical wall making an angle 60° with the ground. How high on the wall does it reach? Solution: Height = 13 * sin60° = 13 * (√3/2) ≈ 11.26 m.
🧮 Formulas
  1. sinθ = opposite/hypotenuse
  2. cosθ = adjacent/hypotenuse
  3. tanθ = opposite/adjacent
  4. cotθ = adjacent/opposite, secθ = hypotenuse/adjacent, cscθ = hypotenuse/opposite
  5. tanθ = sinθ / cosθ
  6. sin(90° − θ) = cosθ, cos(90° − θ) = sinθ, tan(90° − θ) = cotθ
📊 Visual ideas
Right triangle for angle of elevation: horizontal baseline (ground), vertical line (height h), slanted hypotenuse (line of sight) making angle θ at the observer. Label base = d, height = h and show tanθ = h/d. Include a small square at the right angle and an arc marking θ at the base.
Angle of depression diagram: horizontal at observer's eye level, slanted line down to object, vertical from object to ground. Mark the angle of depression at the observer and show the equal angle of elevation at the ground point. Label distances and height; indicate tan relation.
Two‑observer (two triangles sharing the same height): Draw a vertical line for the object height h. From two points on the ground at different distances draw two sight lines to the top forming angles α and β. Label distances d1 and d2. This visual helps write h = d1·tanα and h = d2·tanβ and solve.
Ladder/wall problem: right triangle with hypotenuse = ladder length L, angle with ground θ, vertical reach = L·sinθ, horizontal distance from wall = L·cosθ. Mark L, θ, and show both sin and cos relations.
🔢9

Special Cases and Shortcuts

What this topic covers
In "Some Applications of Trigonometry" (Class 10) the special cases and shortcuts refer to frequently occurring triangle configurations and quick methods that let you solve height-and-distance problems fast without long algebra. The main ideas used are (i) standard trigonometric values for 30°, 45° and 60°, (ii) complementary-angle relations, (iii) angle of elevation = angle of depression symmetry, and (iv) similarity of right triangles.

Key shortcuts and why they work

  • Use standard values directly: tan 30° = 1/√3, tan 45° = 1, tan 60° = √3 (and corresponding sin/cos values). Whenever you get one of these angles, replace trig function by its value immediately to avoid extra algebra.
  • Angle of elevation = angle of depression: If a point above the ground looks down to a point on the ground, the angle of depression from the top equals the angle of elevation from the ground (alternate interior angles). This gives symmetrical right triangles and often lets you set equal distances or heights directly.
  • Complementary-angle shortcut: sin(90° − θ) = cos θ, tan(90° − θ) = cot θ. Use this to convert an unfamiliar function into a familiar one (e.g., tan 60° = cot 30°).
  • Recognise isosceles right triangle (45°–45°–90°): In this case the legs are equal, so if the angle at the base is 45°, height = horizontal distance. That is direct and often solves many ladder/building problems.
  • Similarity-based ratio shortcut: Many problems reduce to similar right triangles. Once you identify similar triangles, set ratios of corresponding sides and eliminate unknowns quickly without solving simultaneous equations explicitly.

Practical step-by-step approach (shortcut recipe)

  1. Draw a clear right triangle marking the angle of elevation or depression and label the horizontal distance (d) and height (h).
  2. If the angle is 30°, 45° or 60°, substitute the known trig value immediately: tanθ = opposite/adjacent → use tanθ value to get h in terms of d (or vice versa).
  3. When two observations are made from different points, look for similar triangles or use difference of distances with tan values; often you can express h in a simple product or ratio of distances and known tan values.
  4. Use angle of depression = angle of elevation to convert a top-down view to a bottom-up view and reuse step (2).

Common mental shortcuts to remember

  • 45° → height = base
  • 30° → height = base × (1/√3)
  • 60° → height = base × √3
  • If you know height and angle 45°, base = height (no calculation needed)
📌 Examples
  • Example 1 — Ladder problem (45°): A ladder leans against a wall making 45° with the ground. If the foot of the ladder is 8 m away from the wall, how high up the wall does it reach? Solution: For 45°, tan 45° = 1 = height / 8, so height = 8 m.
  • Example 2 — Tree and shadow (30°): Sun rays make an angle of elevation of 30° and a tree casts a shadow 5 m long. Find the tree's height. Solution: tan 30° = 1/√3 = height / 5 ⇒ height = 5 × (1/√3) ≈ 2.887 m.
  • Example 3 — Two distances with different angles: From a point A on the ground the angle of elevation to the top of a tower is 60°. Moving 10 m closer to the tower to point B, the angle becomes 75°. Find the tower’s height. (Approach: draw two right triangles with same height h; set h = d1 × tan60 and h = (d1−10) × tan75, equate and solve for d1 then h. Use known tan60 = √3 and tan75 (use calculator or tan(45+30) formula). This reduces algebra because tan60 is a simple exact value.)
  • Example 4 — Angle of depression (plane/runway): A pilot sees the runway at an angle of depression of 5°. If the plane is flying at 1200 m altitude, the horizontal distance to touchdown ≈ 1200 / tan 5°. Use the identity angle of depression = angle of elevation to set up h/ground = tan 5°. This avoids drawing extra non-right triangles.
🧮 Formulas
  1. Fundamental right-triangle relations: sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent.
  2. Special values: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3; sin 45° = √2/2, cos 45° = √2/2, tan 45° = 1; sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3.
  3. Complementary identities: sin(90° − θ) = cos θ, cos(90° − θ) = sin θ, tan(90° − θ) = cot θ.
  4. Angle of elevation = angle of depression (alternate interior angles): use this to convert top-down problems to bottom-up and vice versa.
  5. 45° shortcut: if angle = 45°, then opposite = adjacent (height = horizontal distance).
  6. 30° and 60° shortcuts: if θ = 30°, height = adjacent × (1/√3); if θ = 60°, height = adjacent × √3.
📊 Visual ideas
Unit-circle sketch: show 30°, 45°, 60° points and mark sin and cos coordinates (helps memorize values).
Right-triangle diagram for each special angle: draw three separate right triangles with base 1 and show opposite and hypotenuse for 30°–60°–90° and 45°–45°–90° with numeric side ratios (30° triangle: sides 1, √3, 2; 45° triangle: 1, 1, √2). Label tan = opposite/adjacent on each.
Graph of tan θ vs θ on 0°–90°: mark tan 30° ≈ 0.577, tan 45° = 1, tan 60° ≈ 1.732. Use this to see how tan grows and why small-angle approximations might be used for very small angles.
Angle-of-elevation scenario sketch: ground horizontal line, vertical tower of height h, observer at distance d; draw angle θ at observer and write tan θ = h/d. For two-observer problems draw both observation points on the same baseline to show similar triangles and common height.
🔢10

Use of Calculators, Approximation and Units

Scope: This topic explains how to use calculators correctly in trigonometry, how and when to use approximations (like π ≈ 22/7 or 3.14), handling units of angle (degrees, minutes, seconds and radians), and the effect of rounding/errors. It also gives practical rules to reduce calculation mistakes.

Calculator use — practical rules:

  • Always check the calculator mode: DEG for degrees and RAD for radians. A wrong mode gives incorrect trig values.
  • Use inverse functions with correct domain: if you compute θ = tan-1(value), ensure the value is within the allowed range and mode is correct.
  • Keep extra digits in intermediate steps (use full calculator display) and round only the final answer to required significant figures.

Approximations and rounding:

  • Common approximations: π ≈ 3.14159 (or 22/7 ≈ 3.142857), √2 ≈ 1.414, √3 ≈ 1.732.
  • Significant figures and rounding: keep enough significant figures for the context (usually 3–4 s.f. for answers in Class 10, unless more precision is required). Perform calculations with the calculator's full precision, then round the final result.
  • Absolute error = |approx − exact|. Percent error = (|approx − exact| / exact) × 100%.
  • Small-angle approximations (valid for angle θ in radians and small θ): sin θ ≈ θ, tan θ ≈ θ, cos θ ≈ 1 − θ²/2. Use these for quick estimates and to check plausibility of calculator results.

Units of angle:

  • Degrees: 1 full turn = 360°; minutes and seconds: 1° = 60' (minutes), 1' = 60'' (seconds). Conversion: decimal degrees = degrees + minutes/60 + seconds/3600.
  • Radians: 2π radians = 360°. Conversion between degrees and radians: radians = degrees × (π/180); degrees = radians × (180/π).
  • When graphing or using series/small-angle rules, most calculators and math software expect angles in radians. If you plot sin(x) and use the small-angle approximation sin x ≈ x, interpret x in radians.

Practical tips:

  • For geometry/trigonometry problems, set the calculator to degree mode if angles are given in degrees. If a formula requires radians (e.g., approximations or calculus-based checks), switch to radian mode.
  • When using fractional approximations like 22/7 for π, be aware of the introduced error (22/7 − π ≈ 0.001264). Use 3.1416 or the calculator's π key for better accuracy.
  • Check results for reasonableness: e.g., sin and cos values must lie between −1 and 1; tan values may be large near 90° (π/2 rad).
📌 Examples
  • 1) Angle-mode importance: A vertical pole casts a shadow of 5 m when the angle of elevation of the sun is 40°. Height = shadow × tan(angle) = 5 × tan(40°). Calculator: set DEG mode, tan(40) ≈ 0.8391, height ≈ 5 × 0.8391 = 4.1955 ≈ 4.20 m (rounded to 3 s.f.). If calculator is in RAD, tan(40) gives an absurd value.
  • 2) Degree–radian conversion: Convert 30° to radians: 30 × (π/180) = π/6 ≈ 0.5236 rad. If you need sin(30°) on a radian-mode-only calculator, give 0.5236 as input; sin(0.5236) ≈ 0.5.
  • 3) Small-angle approximation: Estimate sin(0.05 rad). Using the small-angle rule sin θ ≈ θ, sin(0.05) ≈ 0.05. Exact (calculator) sin(0.05) ≈ 0.049979, error ≈ 0.000021 (absolute), percent error ≈ 0.042%. Good accuracy for small θ.
  • 4) Using π ≈ 22/7 error: Approximate circumference of a circle radius 7 cm using π ≈ 22/7: C ≈ 2 × (22/7) × 7 = 44 cm. Exact with π: 2π7 ≈ 43.9823 cm. Absolute error ≈ 0.0177 cm, percent error ≈ 0.0402%. For high precision, use calculator π.
  • 5) DMS to decimal degrees: Convert 12° 34' 18'' to decimal degrees: 12 + 34/60 + 18/3600 = 12 + 0.566666... + 0.005 = 12.571666...° ≈ 12.5717° (rounded).
🧮 Formulas
  1. Degree–radian conversion: radians = degrees × (π/180); degrees = radians × (180/π)
  2. Small-angle approximations (θ in radians, small θ): sin θ ≈ θ, tan θ ≈ θ, cos θ ≈ 1 − θ²/2
  3. Pythagorean identity: sin²θ + cos²θ = 1
  4. Relation: tan θ = sin θ / cos θ
  5. DMS to decimal degrees: decimal° = degrees + minutes/60 + seconds/3600
  6. Absolute error = |approx − exact|; Percent error = (|approx − exact| / exact) × 100%
📊 Visual ideas
Plot y = sin x and y = x on the same axes for x in [−0.5, 0.5] (radians). Label axes with x (radians) and y, and show how sin x ≈ x near 0. Useful to visualize the small-angle approximation.
Plot y = sin x and y = cos x for x in [0, 2π] (or [0°, 360°] if using degree mode). Mark key points (0, π/2, π, 3π/2, 2π) and show range [−1, 1].
Plot y = tan x for x in (−π/2, π/2) and show vertical asymptotes at x = ±π/2. Use radians for accurate shape; annotate that tan grows large near 90°.
Error graph: plot f(x) = sin x − x for x in [−0.5, 0.5] (radians) to show magnitude and sign of small-angle approximation error.

Key Concepts

Angle of elevation
The angle between the horizontal and the line of sight when an observer looks at an object above the horizontal.
Angle of depression
The angle between the horizontal and the line of sight when an observer looks at an object below the horizontal.
Line of sight
The straight line joining the observer's eye to the object being observed.
Horizontal
A straight line parallel to the ground at the observer's eye level; used as the reference for elevation/depression angles.
Vertical
A straight line perpendicular to the horizontal; usually represents the height of an object in height-and-distance problems.
Height
The perpendicular distance from the base to the top of a vertical object (measured along the vertical).
Shadow
The length of the projection of a vertical object on a horizontal surface; relates to the sun's angle of elevation.
Right-angled triangle
A triangle with one 90° angle; the standard model for problems on heights and distances.
Opposite side
In a right triangle, the side opposite to the angle of interest (not the hypotenuse).
Adjacent side
In a right triangle, the side next to the angle of interest (other than the hypotenuse).
Hypotenuse
The side opposite the right angle in a right triangle; the longest side.
Sine (sin)
Trigonometric ratio: sinθ = (opposite side)/(hypotenuse).
Cosine (cos)
Trigonometric ratio: cosθ = (adjacent side)/(hypotenuse).
Tangent (tan)
Trigonometric ratio: tanθ = (opposite side)/(adjacent side) = sinθ/cosθ.
Cotangent (cot)
Reciprocal of tangent: cotθ = 1/tanθ = (adjacent)/(opposite).
Secant (sec)
Reciprocal of cosine: secθ = 1/cosθ = (hypotenuse)/(adjacent).
Cosecant (cosec)
Reciprocal of sine: cosecθ = 1/sinθ = (hypotenuse)/(opposite).
Complementary angles (co-function relation)
Two angles that add to 90°; trigonometric co-function relation: sin(90°−θ)=cosθ, tan(90°−θ)=cotθ, etc.
Inverse trigonometric ratio
Functions arcsin, arccos, arctan that give an angle from a given trigonometric ratio.
Standard angles (30°, 45°, 60°)
Common angles whose sine, cosine and tangent values are memorized and frequently used in problems.

End-of-Chapter Trial Paper & Test Questions

Topic-wise questions to test your understanding of every concept in this chapter.

  1. Define the angle of elevation and the angle of depression. / उन्नयन कोण और अवनमन कोण को परिभाषित कीजिए।
    Show answer

    The angle of elevation is the angle between the horizontal and the line of sight when an observer looks upward at an object; the angle of depression is the angle between the horizontal and the line of sight when the observer looks downward. / उन्नयन कोण वह कोण है जो क्षैतिज और दृष्टि रेखा के बीच बनता है जब प्रेक्षक ऊपर की ओर किसी वस्तु को देखता है; अवनमन कोण वह कोण है जो क्षैतिज और दृष्टि रेखा के बीच बनता है जब प्रेक्षक नीचे की ओर देखता है।

  2. From a point 40 m from the foot of a tower, the angle of elevation of its top is 30°. Find the height of the tower. / एक मीनार के पाद से 40 मीटर दूर एक बिंदु से उसके शिखर का उन्नयन कोण 30° है। मीनार की ऊँचाई ज्ञात कीजिए।
    Show answer

    Using tan 30° = h / 40, we get h = 40 × (1/√3) = 40/√3 ≈ 23.09 m. / tan 30° = h / 40 का प्रयोग करने पर, h = 40 × (1/√3) = 40/√3 ≈ 23.09 मीटर।

  3. Why is the angle of depression of a point from the top of a tower equal to the angle of elevation of the top from that point? / किसी बिंदु का मीनार के शिखर से अवनमन कोण उस बिंदु से शिखर के उन्नयन कोण के बराबर क्यों होता है?
    Show answer

    Because the horizontal line at the top of the tower and the ground are parallel, and the line of sight is a transversal; the two angles are equal alternate interior angles. / क्योंकि मीनार के शिखर पर खींची गई क्षैतिज रेखा और भूमि समांतर हैं, तथा दृष्टि रेखा एक तिर्यक रेखा है; ये दोनों कोण बराबर एकांतर अंतःकोण हैं।

  4. From the top of a lighthouse 50 m high, the angle of depression to a boat is 30°. Find the horizontal distance of the boat from the foot of the lighthouse. / 50 मीटर ऊँचे प्रकाश-स्तंभ के शिखर से एक नाव का अवनमन कोण 30° है। नाव की प्रकाश-स्तंभ के पाद से क्षैतिज दूरी ज्ञात कीजिए।
    Show answer

    The angle of depression equals the angle of elevation from the boat, so tan 30° = 50 / d, giving d = 50 × √3 ≈ 86.60 m. / अवनमन कोण नाव से उन्नयन कोण के बराबर है, इसलिए tan 30° = 50 / d, जिससे d = 50 × √3 ≈ 86.60 मीटर।

  5. A person 1.6 m tall stands 10 m from a tree and finds the angle of elevation of the top of the tree to be 45°. Find the height of the tree. / 1.6 मीटर लंबा एक व्यक्ति एक पेड़ से 10 मीटर दूर खड़ा है और पेड़ के शिखर का उन्नयन कोण 45° पाता है। पेड़ की ऊँचाई ज्ञात कीजिए।
    Show answer

    Height above eye level = 10 × tan 45° = 10 × 1 = 10 m; total tree height = 10 + 1.6 = 11.6 m. / आँख के स्तर से ऊपर की ऊँचाई = 10 × tan 45° = 10 × 1 = 10 मीटर; पेड़ की कुल ऊँचाई = 10 + 1.6 = 11.6 मीटर।

  6. Why is tan used most often (rather than sin or cos) in height-and-distance problems? / ऊँचाई और दूरी की समस्याओं में sin या cos के बजाय tan का प्रयोग सबसे अधिक क्यों किया जाता है?
    Show answer

    Because such problems usually relate the vertical height (opposite side) and the horizontal distance (adjacent side), and tan θ = opposite/adjacent directly connects these two known/unknown quantities without needing the hypotenuse. / क्योंकि ऐसी समस्याएँ प्रायः ऊर्ध्वाधर ऊँचाई (सम्मुख भुजा) और क्षैतिज दूरी (आसन्न भुजा) को जोड़ती हैं, और tan θ = सम्मुख/आसन्न इन दोनों राशियों को कर्ण की आवश्यकता के बिना सीधे जोड़ता है।

  7. From two points on the same straight line towards a tower, the angles of elevation of the top are 60° and 30°, and the distance between the points is 40 m. Find the height of the tower. / एक मीनार की ओर एक ही सीधी रेखा पर स्थित दो बिंदुओं से शिखर के उन्नयन कोण 60° और 30° हैं, तथा बिंदुओं के बीच की दूरी 40 मीटर है। मीनार की ऊँचाई ज्ञात कीजिए।
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    Using h = L·tanα·tanβ/(tanα − tanβ) with α = 60°, β = 30°: tan60·tan30 = 1 and tan60 − tan30 = 2/√3, so h = 40 × 1 ÷ (2/√3) = 20√3 ≈ 34.64 m. / h = L·tanα·tanβ/(tanα − tanβ) में α = 60°, β = 30° रखने पर: tan60·tan30 = 1 तथा tan60 − tan30 = 2/√3, अतः h = 40 × 1 ÷ (2/√3) = 20√3 ≈ 34.64 मीटर।

  8. If a vertical pole casts a shadow equal in length to its own height, what is the angle of elevation of the sun? Justify. / यदि एक ऊर्ध्वाधर खंभे की छाया की लंबाई उसकी अपनी ऊँचाई के बराबर है, तो सूर्य का उन्नयन कोण क्या है? औचित्य दीजिए।
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    Here tan θ = height/shadow = 1, so θ = 45°; the height equals the shadow only when tan θ = 1, which occurs at 45°. / यहाँ tan θ = ऊँचाई/छाया = 1, इसलिए θ = 45°; ऊँचाई छाया के बराबर तभी होती है जब tan θ = 1, जो 45° पर होता है।

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