Class 10 Mathematics Unit 15 – Trigonometry | Nepal eNotes

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CLASS 10 MATHEMATICS • UNIT 15

Unit 15: Trigonometry

Trigonometric ratios, angles of elevation and depression, and practical height-and-distance problems — reconstructed from the supplied 33-page handwritten notes.

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Source note: The uploaded filename says “Unit 15 – Circle”, but page 1 of the handwritten PDF is headed “Lesson 15: Trigonometry”. The scanned page content is therefore used as the source of truth. Where a handwritten measurement or question is unclear, the source method/result is identified without inventing missing values.

PDF Page 1

Trigonometric Ratios

For a right-angled triangle, with respect to angle θ:

Sine

sin θ = p / h

Cosine

cos θ = b / h

Tangent

tan θ = p / b

Cosecant

cosec θ = h / p

Secant

sec θ = h / b

Cotangent

cot θ = b / p

p = perpendicular, b = base, and h = hypotenuse.

PDF Page 1

Standard Trigonometric Values

θ	0°	30°	45°	60°	90°
sin θ	0	1/2	1/√2	√3/2	1
cos θ	1	√3/2	1/√2	1/2	0
tan θ	0	1/√3	1	√3	∞

PDF Pages 1–2

Angle of Elevation and Angle of Depression

Angle of Elevation

When an observer sees an object above the horizontal level of the observer’s eyes, the angle formed between the line of sight and the horizontal line is called the angle of elevation.

Angle of Depression

When an observer sees an object below the horizontal level of the observer’s eyes, the angle formed between the line of sight and the horizontal line is called the angle of depression.

PDF Pages 2–33

Exercise 15

PDF Pages 2–4

2. Find the Value of the Unknown in the Right-Angled Triangles

(a) Perpendicular = 5√3 m, angle = 60°, base = n

tan 60° = p / b

√3 = 5√3 / n

n = 5√3 / √3

∴ n = 5 m

(b) Base = 5√3 m, angle = 30°, perpendicular = m

tan 30° = p / b

1/√3 = m / (5√3)

∴ m = 5 m

(c) Perpendicular = 10 m, base = 10 m, angle = n

tan n = 10 / 10 = 1

tan n = tan 45°

∴ n = 45°

PDF Pages 4–6

3. Tower Height and Distance Problems

(a) Tower height = 60 m, angle of elevation = 30°. Find the horizontal distance.

tan 30° = 60 / n

1/√3 = 60/n

∴ n = 60√3 m

(b) Tower height = 12 m and horizontal distance = 12 m. Find the angle of elevation.

tan n = 12/12 = 1

tan n = tan 45°

∴ n = 45°

(c) Horizontal distance = 12 m and angle of elevation = 45°. Find the tower height.

tan 45° = n/12

1 = n/12

∴ n = 12 m

PDF Pages 6–9

4. Broken Tree Problems

These questions use the fact that the broken upper part of the tree becomes the hypotenuse of a right triangle, while the remaining upright portion is the perpendicular.

(a) Tree of total height 10 m, broken so that the top touches the ground and makes 60° with the ground.

The source sets up:

sin 60° = (remaining vertical part) / (broken part)

and uses the relation:

remaining part + broken part = 10 m

(b) Remaining upright part = 7.5 m, angle with the ground = 60°

sin 60° = 7.5 / n

√3/2 = 7.5/n

n = 15/√3 = 5√3 m

Total length of tree before breaking:

= 7.5 + 5√3

≈ 16.16 m

(c) Broken part = 30 m, angle with the ground = 30°

sin 30° = n/30

1/2 = n/30

∴ n = 15 m

Total tree length before breaking:

= 30 + 15

∴ 45 m

Horizontal distance:

BC = √(30² − 15²)

= √675

= 15√3 m

Part (a) is less legible than parts (b) and (c). The source setup is preserved, but the final handwritten value is not clear enough to state confidently.

PDF Pages 9–12

5. Heights of Towers, Kites and Trees

(a) A 1.7 m tall man stands 30 m from a tower; angle of elevation = 60°.

tan 60° = n/30

√3 = n/30

n = 30√3

Total height of tower = 30√3 + 1.7

≈ 53.66 m

(b) A 2 m tall man flies a kite from the roof of a 33.6 m high house. String length = 90√2 m and angle = 45°.

Height of the man’s hand/eye above ground = 33.6 + 2 = 35.6 m

sin 45° = (m − 35.6)/(90√2)

1/√2 = (m − 35.6)/(90√2)

90 = m − 35.6

∴ m = 125.6 m

(c) A 1.5 m tall man observes the top of a 51.5 m high tree at 45°. Find the distance.

Vertical difference = 51.5 − 1.5 = 50 m

tan 45° = 50/n

1 = 50/n

∴ n = 50 m

PDF Pages 12–15

6. Tower and House Problems

(a) Horizontal distance = 20 m and angle of elevation = 60°

The source forms a right triangle and uses:

tan 60° = perpendicular / 20

The handwritten calculation then compares this vertical difference with a given height of 36.5 m.

(b) From the top of a 30 m high house, the angle of depression to the base of a tree is 30°; horizontal distance = 10√3 m.

tan 30° = (30 − n)/(10√3)

1/√3 = (30 − n)/(10√3)

10 = 30 − n

∴ n = 20 m

(c) Tower height = 60 m, distance between tower and house = 35 m, angle of depression to top of house = 45°.

tan 45° = (60 − n)/35

1 = (60 − n)/35

35 = 60 − n

∴ height of house = 25 m

The wording of part (a) is partly unclear in the scan. Its right-triangle setup is retained without forcing a reconstructed question statement.

PDF Pages 15–18

7. Circular Pond and Pole

(a) Diameter of pond = 90 m; pole at centre; angle of elevation from edge = 45°.

Radius = 90/2 = 45 m

tan 45° = pole height / 45

1 = pole height / 45

∴ pole height = 45 m

(b) Diameter of pond = 130 m; angle of elevation = 45°.

Radius = 130/2 = 65 m

tan 45° = pole height / 65

∴ pole height = 65 m

(c) Pole height = 11.62 m above the water; observer height = 1.62 m; angle of elevation = 30°.

Vertical difference = 11.62 − 1.62 = 10 m

tan 30° = 10/BE

1/√3 = 10/BE

BE = 10√3 m

Diameter of pond = 2 × BE

∴ diameter = 20√3 m

PDF Pages 18–21

8. Kite Problems

(a) Kite string = 120 m, angle with horizontal = 30°, observer height = 1.5 m.

sin 30° = m/120

1/2 = m/120

m = 60 m

Height of kite from ground = 60 + 1.5

∴ 61.5 m

(b) Kite flown from the roof of a house

Pages 19–20 contain another kite problem involving a 1.5 m tall person, a house roof, a 30° angle and a given string length. The source uses the sine ratio to find the vertical rise and then adds the house and observer heights.

(c) Man height = 2 m, house height = 32 m, string = 66√2 m, angle = 60°.

sin 60° = n/(66√2)

√3/2 = n/(66√2)

2n = 66√6

n = 33√6

Height from ground = 33√6 + 32 + 2

≈ 114.83 m

The measurements in part (b) are blurred/overwritten. The visible source method is preserved, but uncertain values are not invented.

PDF Pages 21–23

9. Shadow Problems

Pages 21–23 compare the height of a tower/pole with the length of its shadow at different times of day.

First case

Height = 20 m, shadow = 20√3 m

tan θ = 20/(20√3) = 1/√3

∴ θ = 30°

Second case

The source again uses tan 30° = height/shadow and calculates a shadow length of:

∴ 75 m

The full sentence of Question 9 is partly faint, but the two trigonometric calculations above are clearly visible in the handwritten solution.

PDF Pages 23–24

10. Tree at One Corner of a School

A tree is 25 m high. A 1.2 m tall man stands at another corner, at a distance of 23.8 m from the tree.

Vertical difference = 25 − 1.2 = 23.8 m

tan θ = 23.8/23.8 = 1

tan θ = tan 45°

∴ θ = 45°

PDF Pages 24–26

12. Circular Pond Application

Circumference of the circular pond = 176 m. A pole is fixed at the centre. A 1.6 m tall man at the edge observes the top of the pole at 45°.

(a) Find the distance between the man and the pole.

C = 2πr

176 = 2 × 22/7 × r

∴ r = 28 m

Distance between man and pole = 28 m

(b) Find the height of the pole above the water surface.

tan 45° = n/28

n = 28 m

Total pole height above water = 28 + 1.6

∴ 29.6 m

(c) By how much should the pole be shortened so that the angle becomes 30°?

tan 30° = n/28

1/√3 = n/28

n = 28/√3 ≈ 16.1658 m

Required reduction in vertical height:

28 − 16.1658 ≈ 11.8342 m

PDF Pages 26–28

13. Kite from the Roof of a House

A 1.2 m tall man flies a kite from the roof of an 8.8 m high house. The string is 180 m long and makes an angle of 30° with the horizontal.

sin 30° = BC/180

1/2 = BC/180

BC = 90 m

Total height from ground:

= 1.2 + 8.8 + 90

∴ 100 m

PDF Pages 28–30

14. Two Buildings

Two buildings have heights 20 m and 32 m, and the distance between them is 12 m.

(a) Angle from the roof of the smaller building to the roof of the taller building

Difference in height = 32 − 20 = 12 m

tan θ = 12/12 = 1

∴ θ = 45°

(b) Angle of elevation

The source explains that the angle formed between the horizontal line through the observer and the upward line of sight is the angle of elevation.

(c) Angle of depression

By the parallel-horizontal-line relationship, the angle of depression from the taller building is equal to the corresponding angle of elevation:

∴ 45°

(d) Ladder/string joining the two roofs

sin 45° = 12/m

1/√2 = 12/m

∴ m = 12√2 m

PDF Pages 30–31

15. Tower and House on the Same Level

The source gives a tower and a house 25 m apart, with an angle of elevation of 45° from the top of the house to the top of the tower.

tan 45° = n/25

n = 25 m

The source then adds the house height to this vertical difference and records the tower height as:

∴ 40 m

The handwritten house height appears as 15 m in the working, while the question line is slightly blurred. The source’s final total of 40 m is retained as written.

PDF Pages 31–33

16. Angles of Depression from the Top of a 50 m Tower

Pages 31–33 show a 50 m high tower and a nearby house. Two downward lines of sight are drawn from the top of the tower to points on the house/ground.

House-height calculation shown in the source

Horizontal distance = 20 m

tan 60° = n/20

n = 20√3

Height of house = 50 − 20√3

≈ 15.36 m

Another distance calculation

The source also uses:

tan 45° = n/20

n = 20 m

The exact wording of the sub-parts on pages 31–33 is partly blurred. The numerical relations and final values above are the clearly readable parts of the handwritten solution.