What is the main concept of 'Some Applications of Trigonometry'?

This chapter explains how trigonometric ratios help find the height or distance of an object using angles of elevation or depression.

Where is trigonometry used in real life?

Trigonometry is used in architecture, astronomy, navigation, and engineering for accurate measurement of heights and distances.

Are RAAM SETU notes helpful for competitive exams?

Yes, these notes are useful for NEET, SSC, Railway, and Police Bharti exams as they strengthen maths fundamentals and problem-solving.

Some Applications of Trigonometry — Class 10 Maths Notes

📘 Introduction

In this chapter, we explore how trigonometric ratios like sine, cosine, and tangent help us calculate heights and distances of real-world objects such as towers, mountains, and airplanes. These applications make trigonometry one of the most practical topics in mathematics — essential for NEET, SSC, Railway, and Police Bharti aspirants.

🔹 Key Concepts & Formulas

Angle of Elevation — Angle formed by the line of sight above the horizontal line.
Angle of Depression — Angle formed by the line of sight below the horizontal line.
tan θ = Opposite / Adjacent
sin θ = Opposite / Hypotenuse
cos θ = Adjacent / Hypotenuse

📊 Chapter Summary Table

Topic	Key Formula	Use Case
Angle of Elevation	tan θ = h/d	Find height from distance
Angle of Depression	tan θ = h/d	Find distance from height
Multiple Angles	Use two triangles	Complex height problems

🧮 Practice Questions

A tower is 50 m high. The angle of elevation of its top from a point on the ground is 30°. Find the distance of the point from the tower’s base.
The angle of elevation of the top of a tree from a point 36 m away from its base is 45°. Find the height of the tree.
From a building top 60 m high, the angle of depression to a car on the road is 30°. Find the distance of the car from the building.

1. Quick Recap of Trig Ratios

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

Use these as building blocks — almost every applied problem reduces to one right triangle at a time.

2. Angle of Elevation & Angle of Depression

Angle of elevation — angle between line of sight upwards and horizontal. Angle of depression — angle between line of sight downwards and horizontal.

In problems, angles of elevation/depression from the same horizontal are equal (alternate interior angles). Always draw the horizontal through the observer.

3. Single-step Heights & Distances (One station)

Standard pattern: distance on ground (d), height h, angle θ. Use either tan or sin/cos depending on knowns.

Template: If angle of elevation = θ and horizontal distance = d, then h = d · tan θ.

Or, if slant distance (hypotenuse) s is known: h = s · sin θ.

4. Two-station Problems (Two Observers or Two Angles)

These problems give angles from two points at known separation. Method:

Place coordinate: foot of object at C; observers at A and B on same horizontal line with AB known.
From tan relations: h = x·tan α and h = (x + AB)·tan β (if A nearer B further). Solve for x then get h.

Often leads to linear equation in x: x·tan α = (x + d)·tan β ⇒ x(tan α − tan β) = d·tan β ⇒ x = d·tan β/(tan α − tan β).

5. Bearings & Navigation

Bearing is measured clockwise from North. For example, bearing 045° means 45° east of north. In navigation problems, convert bearings to angle with horizontal or between lines to apply trigonometry.

Convert bearings to standard angle measures and draw clear north-south and east-west references. Use vector-like decomposition into perpendicular components and apply trig.

6. Real-world Applications

Surveying: measuring inaccessible heights (towers, cliffs) and distances using angle observations.
Navigation: course plotting, bearing calculations and triangulation.
Architecture: determining slopes, roof pitches and sight-lines.
Engineering: resolving forces into components, incline problems.

7. Solved Examples

Example 1 (Single station). Angle of elevation of top of tower from point A is 35°. If A is 40 m from foot of tower, find height (use tan 35° ≈ 0.7002).

h = 40 × tan 35° ≈ 40 × 0.7002 ≈ 28.01 m.

Example 2 (Two-station). From two points A and B on same level, separated by 30 m, angles of elevation to top of tower are 45° and 30° respectively (A closer). Find height.

Let distance from A to foot = x. Then h = x·tan45 = x and h = (x + 30)·tan30 = (x + 30)/√3.
So x = (x + 30)/√3 ⇒ x(1 − 1/√3) = 30/√3 ⇒ x = [30/√3]/(1 − 1/√3). Solve numerically to get h = x.

Example 3 (Bearing / Navigation). Ship P is at 10 km due south of Q. Bearing of ship R from P is 045°, and bearing of R from Q is 315°. Find distance PR and QR and coordinates if needed (use decomposition).

Approach: place Q at origin, Q at (0,0), P at (0,−10). Convert bearings to direction vectors and solve intersection or use trigonometric relations to get distances.

8. Practice Questions

From a point 20 m away from a tower, angle of elevation is 60°. Find height of tower.
Two points A and B are 50 m apart. Angles of elevation to top of a building from A and B are 30° and 45° respectively. Find height.
A ship sails from a lighthouse. From point A (on shore) bearing to lighthouse is 060°, from point B (30 km east of A) bearing is 330°. Find approximate position of lighthouse (use basic trig and coordinate assumptions).
Prove that in two-station problems, if tan α = tan β then no unique height exists (explain geometrically).

9. Exam Tips & Strategy

Always draw accurate diagrams and mark distances/angles clearly.
Do unit-checks (meters, km) and choose consistent units.
In two-station problems, keep track of which observer is closer — it affects sign in formulas.
Use calculators for tangent values; memorise tan 30°, tan45°, tan60° for quick checks.