Class 10 Applications of Trigonometry Notes

Exam Weightage & Blueprint

Total: ~4-6 Marks

This chapter falls under Unit V: Trigonometry (12 marks total). As per the latest syllabus: Simple problems on heights and distances. Problems should not involve more than two right triangles. Angles of elevation / depression should be only 30°, 45°, 60°.

Question Type	Marks	Frequency	Focus Topic
MCQ	1	Low	Simple height finding ($\tan \theta$)
Short Answer	3	Medium	Single Triangle Problems (Broken Tree, etc.)
Long Answer / Case Study	4-5	Very High	Double Triangle Problems (Two ships, Building & Tower)

⏰ Last 24-Hour Checklist

Angle of Elevation: Looking UP from horizontal.
Angle of Depression: Looking DOWN from horizontal.
Key Relation: Angle of Depression = Angle of Elevation (Alternate Angles).

Values: Memorize $\tan 30^\circ, \tan 45^\circ, \tan 60^\circ$.
Diagram Rule: Always mark the right angle and the reference angle.
Root Values: $\sqrt{3} = 1.732$ (Use only if asked).

🔭 Core Theory & Definitions

Line of Sight: The line drawn from the eye of an observer to the point in the object viewed by the observer.

Angles Explained

Angle of Elevation

The angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level.

Angle of Depression

The angle formed by the line of sight with the horizontal when the point being viewed is below the horizontal level.

Important: In Angle of Depression problems, always draw a horizontal line from the observer's eye level first. Then mark the angle downwards.

📐 Essential Formulas & Values

Most questions are solved using the tangent ratio:

$$ \tan \theta = \frac{\text{Perpendicular (Height)}}{\text{Base (Distance)}} $$

Values to Remember

30-
$\tan 30^\circ = \frac{1}{\sqrt{3}}$

45-
$\tan 45^\circ = 1$

60-
$\tan 60^\circ = \sqrt{3}$

Solved Examples (Board Marking Scheme)

Q1. A tower stands vertically on the ground. From a point 15 m away from the foot, the angle of elevation is $60^\circ$. Find the height. (2 Marks)

Step 1: Diagram & Given 0.5 Mark

Let AB be the tower ($h$). Point C is 15 m away. $\angle ACB = 60^\circ$.

Step 2: Apply Ratio 1 Mark

In $\triangle ABC$, $\tan 60^\circ = \frac{AB}{BC} = \frac{h}{15}$.

Step 3: Solve 0.5 Mark

$\sqrt{3} = \frac{h}{15} \implies h = 15\sqrt{3}$ m.

Q2. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is $60^\circ$ and the angle of depression of its foot is $45^\circ$. Determine the height of the tower. (5 Marks)

Step 1: Setup & Diagram 1 Mark

Let AB be the building (7m) and CD be the tower ($H$). Draw horizontal line AE.

$\angle CAE = 60^\circ$ (Elevation), $\angle EAD = 45^\circ$ (Depression).

Step 2: Triangle 1 (Lower) 1.5 Marks

In $\triangle ABD$, $\tan 45^\circ = \frac{AB}{BD} = \frac{7}{BD}$.

$1 = \frac{7}{BD} \implies BD = 7$ m.

Also, $AE = BD = 7$ m.

Step 3: Triangle 2 (Upper) 1.5 Marks

In $\triangle AEC$, $\tan 60^\circ = \frac{CE}{AE}$.

$\sqrt{3} = \frac{CE}{7} \implies CE = 7\sqrt{3}$ m.

Step 4: Total Height 1 Mark

Height of tower = $CE + ED = 7\sqrt{3} + 7 = 7(\sqrt{3} + 1)$ m.

Previous Year Questions (PYQs)

2023 (Case Study): A kite is flying at a height of 60m. The string makes an angle of $60^\circ$ with the ground. Find string length.
Ans: Use $\sin 60^\circ = \frac{P}{H} \implies \frac{\sqrt{3}}{2} = \frac{60}{L} \implies L = \frac{120}{\sqrt{3}} = 40\sqrt{3}$ m.

2020 (3 Marks): The shadow of a tower is $40$m longer when the Sun's altitude is $30^\circ$ than when it is $60^\circ$. Find the height.
Ans: $h(\cot 30^\circ - \cot 60^\circ) = 40 \implies h(\sqrt{3} - \frac{1}{\sqrt{3}}) = 40 \implies h = 20\sqrt{3}$ m.

2019 (4 Marks): Two poles of equal heights stand on either side of an 80m wide road. Angles of elevation are $60^\circ$ and $30^\circ$. Find height and distances.
Ans: $h = 20\sqrt{3}$ m. Distances are 20m and 60m.

Exam Strategy & Mistake Bank

⚠️ Mistake Bank

Wrong Ratio: Using $\tan$ when hypotenuse (ladder/kite string) is involved. Use $\sin$ or $\cos$ instead.

Angle Place: Marking the angle of depression with the vertical instead of the horizontal.

Calculation: $\sqrt{3} \approx 1.732$. Use this value only if specified in the question.

💡 Scoring Tips

Diagram is Mandatory: Even if you solve it correctly, no marks for diagram = marks deducted.

Label Vertices: Always label the triangle vertices (A, B, C) and state "Let AB be the tower...".

Double Triangles: In 5-mark questions, you usually get two equations with 'h' and 'x'. Isolate 'x' in one and substitute in the other.

📝 More Solved Board Questions

Q3. A 1.5m tall boy is standing at some distance from a 30m tall building. The angle of elevation from his eyes to the top of the building increases from $30^\circ$ to $60^\circ$ as he walks towards the building. Find the distance he walked. 4 Marks

Sol. Height above eye level = $30 - 1.5 = 28.5$ m.

In $\triangle 1$ ($60^\circ$): $\tan 60^\circ = \frac{28.5}{x} \implies x = \frac{28.5}{\sqrt{3}}$ m.

In $\triangle 2$ ($30^\circ$): $\tan 30^\circ = \frac{28.5}{x+d} \implies \frac{1}{\sqrt{3}} = \frac{28.5}{x+d} \implies x+d = 28.5\sqrt{3}$.

$d = 28.5\sqrt{3} - \frac{28.5}{\sqrt{3}} = \frac{28.5(3-1)}{\sqrt{3}} = \frac{57}{\sqrt{3}} = 19\sqrt{3}$ m.

Answer: $19\sqrt{3}$ m

🎯 Board Pattern (2018–2025): Case Study questions in Chapter 9 often involve a moving object (car/boat) or a change in elevation (like the boy walking towards the building). Always find the horizontal distances in both scenarios and subtract/add them.

📋 Board Revision Checklist

✅ Diagram check: Right angle marked? Observer level marked?
✅ Elevation = Looking UP; Depression = Looking DOWN
✅ If observer height is given (e.g., 1.5m boy), subtract it from the total object height
✅ $\tan 30^\circ = 1/\sqrt{3} \approx 0.577$
✅ $\tan 45^\circ = 1$ (Shadow length = Object height)
✅ $\tan 60^\circ = \sqrt{3} \approx 1.732$
✅ For "broken tree" problems: Total height = Standing part + Hypotenuse (Broken part)
✅ Alternate angles: Depression from top = Elevation from ground

💡 Exam Tip:
In double-triangle problems, always solve the triangle with more known information first (usually the one with $45^\circ$ angle as it makes Base = Perpendicular).

Concept Mastery Quiz 🎯

Test your readiness for the board exam.

1. The length of a shadow of a tower is $\sqrt{3}$ times its height. The angle of elevation of the sun is:

2. A bridge over a river makes an angle of $45^\circ$ with the river bank. If the length of the bridge across the river is 150m, the width of the river is:

3. The angle of depression of a car parked on the road from the top of a 150m high tower is $30^\circ$. The distance of the car from the tower is:

4. If a pole 6m high casts a shadow $2\sqrt{3}$m long on the ground, then the sun's elevation is:

5. A ladder 10m long reaches a window 8m above the ground. The distance of the foot of the ladder from the base of the wall is:

Chapter 9: Applications of Trigonometry

Overview