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CBSE Class 10 · Maths · board exam 2027

Chapter 9: Some Applications of Trigonometry — notes

The chapter’s 6 parts in the book’s order — the first idea of each and the points Medha Ma’am writes on the board — and the summary at the end. The full lesson and a check question for each part are in the app.

This chapter’s important questions →

  1. 1. Where is trigonometry used?

    Look at this figure.

    • • Astronomy: distances to the planets and the stars
    • • Geography and navigation: maps, the position of an island
    • • Surveying: theodolite (measures angles with a rotating telescope)
    • • Great Trigonometric Survey → the highest mountain → named after Sir George Everest

  2. 2. Line of sight, angle of elevation and angle of depression

    The figure.

    • • Line of sight: the line from the eye to the point being viewed
    • • Horizontal line: the level line through the eye, parallel to the ground
    • • Angle of elevation: the object is above, we raise our head
    • • Angle of elevation: between the line of sight and the horizontal (above the horizontal)
    • • Angle of depression: the object is below, we lower our head
    • • Angle of depression: between the line of sight and the horizontal (below the horizontal)
    • • Both angles: measured from the horizontal line, not from the vertical line
    • • Angle of depression of C from A = 30°
    • • Horizontal line through A ∥ ground, AC is a transversal
    • • Alternate angles are equal ⇒ angle of elevation of A from C = 30°

  3. 3. Height of a tower from one right triangle

    Every question starts with a figure.

    • • AB: the tower (A top, B foot), ∠B = 90°
    • • BC: distance from the foot, ∠ACB: angle of elevation
    • • tan C = perpendicular/base = AB/BC
    • • Height and distance → tan (or cot)
    • • Height and slanting length (ladder, string) → sin
    • • Distance and slanting length → cos
    • • Example: BC = 15 m, ∠ACB = 60°, AB = ?
    • • tan 60° = AB/BC
    • • √3 = AB/15
    • • AB = 15 × √3 = 15√3
    • • Height of the tower = 15√3 m
    • • √3 ≈ 1.732
    • • 15 × 1.732 = 25.98
    • • AB ≈ 25.98 m
    • • BC = 20 m ⇒ √3 = AB/20
    • • AB = 20√3 m
    • • Full height of the tower = part above eye level + height of the observer

  4. 4. Ladder and chimney: sin, cot and the observer's height

    An electrician's problem.

    • • Pole AD = 5 m (A top, D foot), AB = 1.3 m
    • • Ladder BC, C on the ground, ∠BCD = 60°, ∠D = 90°
    • • BD = AD − AB = 5 − 1.3 = 3.7
    • • BD = 3.7 m
    • • sin 60° = BD/BC
    • • √3/2 = 3.7/BC
    • • BC × √3 = 3.7 × 2 = 7.4
    • • BC = 7.4/√3 ≈ 7.4/1.73 ≈ 4.28 m
    • • cot 60° = DC/BD
    • • 1/√3 = DC/3.7
    • • DC = 3.7/√3 ≈ 3.7/1.73 ≈ 2.14 m
    • • Chimney AB, eye D, E on AB at eye level
    • • DE = 28.5 m, BE = 1.5 m, ∠ADE = 45°
    • • tan 45° = AE/DE
    • • 1 = AE/28.5 ⇒ AE = 28.5 m
    • • AB = AE + BE = 28.5 + 1.5 = 30
    • • Height of the chimney = 30 m
    • • AB = 28.5 + 2 = 30.5
    • • Height of the chimney = 30.5 m

  5. 5. Two triangles, one common side

    Questions in which two right triangles are formed.

    • • Building AB = 10 m (A foot, B top), flagstaff BD = x m
    • • ∠APB = 30°, ∠APD = 45°, ∠A = 90°
    • • In △PAB: tan 30° = AB/AP
    • • 1/√3 = 10/AP
    • • AP = 10√3 m ≈ 17.32 m
    • • In △PAD: AD = 10 + x
    • • tan 45° = AD/AP
    • • 1 = (10 + x)/10√3
    • • 10 + x = 10√3
    • • x = 10√3 − 10 = 10(√3 − 1)
    • • x = 10(1.732 − 1) = 10 × 0.732 = 7.32
    • • Length of the flagstaff = 7.32 m
    • • The common side AP joined the two triangles
    • • Tower AB = h, BC = x (shadow at 60°), BD = x + 40 (shadow at 30°)
    • • CD = 40 m
    • • tan 60° = h/x ⇒ √3 = h/x ⇒ h = x√3 … (1)
    • • tan 30° = h/(x + 40) ⇒ 1/√3 = h/(x + 40) ⇒ h = (x + 40)/√3 … (2)
    • • x√3 = (x + 40)/√3
    • • 3x = x + 40
    • • 2x = 40 ⇒ x = 20
    • • h = 20√3 m

  6. 6. Questions on the angle of depression

    The angle of depression.

    • • Multi-storeyed building PC (C foot), small building AB = 8 m (B top, A foot)
    • • Draw BD ⊥ PC, so DC = AB = 8 m and BD = AC
    • • Alternate angles ⇒ ∠PBD = 30°, ∠PAC = 45°
    • • tan 45° = PC/AC = 1 ⇒ AC = PC
    • • PC = PD + DC = PD + 8
    • • BD = AC = PD + 8 … (1)
    • • tan 30° = PD/BD ⇒ BD = PD√3 … (2)
    • • PD√3 = PD + 8
    • • PD(√3 − 1) = 8
    • • PD = 8/(√3 − 1) × (√3 + 1)/(√3 + 1)
    • • = (8(√3 + 1))/(3 − 1) = (8(√3 + 1))/2 = 4(√3 + 1)
    • • PD = 4(√3 + 1) m
    • • PC = 4(√3 + 1) + 8 = 4√3 + 12 = 4(3 + √3)
    • • Height PC = distance AC = 4(3 + √3) m
    • • PD = 3 m, ∠PAD = 30°, ∠PBD = 45°, AB = AD + DB
    • • 1/√3 = 3/AD ⇒ AD = 3√3 m
    • • 1 = 3/BD ⇒ BD = 3 m
    • • AB = 3√3 + 3 = 3(√3 + 1)
    • • AB ≈ 3(1.73 + 1) = 3 × 2.73 = 8.19
    • • Width of the river = 3(√3 + 1) m ≈ 8.19 m

Summary

Very good! First we understood the line of sight and the horizontal line. If the object is above, it is the angle of elevation, and if it is below, it is the angle of depression. Both angles are measured from the horizontal line, and because they are alternate angles, the angle of depression from the top equals the angle of elevation from below. Then we solved every question the same way. First the figure, then let, then the right ratio. Tan for height and distance, sin or cos when there is a ladder or a string. When two triangles were formed, we joined the two equations through their common side. And at the end we wrote the answer with its unit. Write today's three rules once in your notebook without looking, and always draw the figure before every question. Today's homework is on your screen. If anything was not clear, press the Ask button and ask me. Done, you worked really well today!

AskMedha · Medha Ma’am — CBSE Class 10 Maths, Chapter 9. Written from the textbook, checked twice.

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