CBSE Class 10 Maths Chapter 9 Notes — Some Applications of Trigonometry

Key Definitions

Line of sight: The line drawn from the eye of the observer to the object being viewed.

Angle of Elevation: The angle between the horizontal and the line of sight upward to the object.

• Measured from horizontal up to the line of sight

Angle of Depression: The angle between the horizontal and the line of sight downward to the object.

• Measured from horizontal down to the line of sight

Important relationship: Angle of elevation from A to B = Angle of depression from B to A (alternate interior angles with horizontal lines as parallels)

Setting Up Problems — Method

1. Draw a clear diagram — this is the most important step 2. Label all given distances and angles 3. Mark the right angle (usually where the vertical meets the horizontal) 4. Identify which trigonometric ratio to use:

5. Write the trig equation and solve

• If you know the angle and hypotenuse → use sin or cos
• If you know the angle and one leg → use tan

Standard Results to Recognise

When angle = 30°: tan 30° = 1/√3 → opposite = adjacent/√3 When angle = 45°: tan 45° = 1 → opposite = adjacent (isoceles right triangle) When angle = 60°: tan 60° = √3 → opposite = √3 × adjacent

Worked Examples — Single Angle

Type 1: Find height of a tower

A tower casts a shadow 30 m long when the Sun's elevation is 30°. Find the height.

Draw: Tower = AB (vertical), Shadow = BC (horizontal), angle at C = 30°

tan 30° = AB/BC → 1/√3 = h/30 → h = 30/√3 = 10√3 m

Type 2: Find distance from an observer

The angle of elevation of the top of a 20 m tall building from a point on the ground is 60°. Find the distance from the point to the foot of the building.

tan 60° = 20/d → √3 = 20/d → d = 20/√3 = 20√3/3 m

Type 3: Angle of depression

From the top of a 12 m high building, the angle of depression of a car is 45°. How far is the car from the building?

Let building height = AB = 12 m. Car at C on ground. Angle of depression from A = 45°. Angle of elevation from C = 45° (alternate angles) tan 45° = AB/BC → 1 = 12/BC → BC = 12 m

Worked Examples — Two Angles (Two Buildings / Different Positions)

Type 4: Two buildings/towers

Two poles of equal heights are standing opposite each other on either side of a road 80 m wide. From a point between them on the road, the angles of elevation are 60° and 30°. Find the heights and position of the point.

Let height = h, point divides 80 m as x and (80−x).

From pole 1: tan 60° = h/x → h = x√3 ... (i) From pole 2: tan 30° = h/(80−x) → h = (80−x)/√3 ... (ii)

Equate: x√3 = (80−x)/√3 → 3x = 80−x → 4x = 80 → x = 20 m h = 20√3 m → h = 20√3 m

Point is 20 m from pole 1, 60 m from pole 2.

Type 5: Observer moving toward tower

A man observes the top of a tower of height h at angle of elevation 30°, then walks d metres closer and observes it at 60°. Prove h = d√3/2.

Initial position: tan 30° = h/(x+d) → x+d = h√3 ... (i) Final position: tan 60° = h/x → x = h/√3 ... (ii)

From (i) − (ii): d = h√3 − h/√3 = h(3−1)/√3 = 2h/√3

∴ h = d√3/2 ✓

Type 6: Height of a flying kite

A kite is flying at 75 m above the ground. The string makes a 60° angle with the ground. Find the length of the string.

sin 60° = 75/L → √3/2 = 75/L → L = 150/√3 = 50√3 m

Quick Reference — Which Formula?

Board Exam Tips

• 5–6 marks from this chapter (usually one 5-mark problem)
• Always draw the diagram first — most errors come from wrong diagram, not wrong calculation
• This chapter is pure application — no new trigonometry, just tan/sin/cos from Chapter 8
• The 5-mark problems usually involve two angles or two towers
• Show the equation clearly: tan 30° = h/d → then substitute value of tan 30°
• Leave answers in surd form (e.g., 10√3) unless decimal is asked
• "Angle of depression = angle of elevation (alternate angles)" — mention this when converting depression to elevation in your working

📚 RD Sharma Mathematics Class 10 — Height and Distance Problems

📝 Oswaal CBSE Maths Question Bank — Applications of Trigonometry