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How can a surveyor measure the height of a 500-meter radio tower or the distance of a ship from a lighthouse without climbing the tower or getting wet...
How can a surveyor measure the height of a 500-meter radio tower or the distance of a ship from a lighthouse without climbing the tower or getting wet? Heights and Distances apply trigonometry to solve real-world problems.
Why This Chapter Matters
In Class 10 Mathematics, "Some Applications of Trigonometry" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.
Before You Begin (Prerequisites)
- Trigonometric values ($30^\circ, 45^\circ, 60^\circ$) from Chapter 8.
- Right triangles.
- Angle measurement.
What You Will Learn (Core Objectives)
- Define Line of Sight, Horizontal Line, Angle of Elevation, and Angle of Depression.
- Draw accurate geometric right-triangle diagrams from verbal word descriptions.
- Calculate heights of trees, towers, and buildings using tangent ratios.
- Calculate distances between moving ships or cars observed from a lighthouse.
- Solve dual-triangle problems with two angles of elevation/depression.
Chapter Roadmap & Progression
1
1. Line of Sight, Elevation & Depre...
2
2. The Tangent Ratio Method
Complete Concept Guide (100% Curriculum Coverage)
1. Line of Sight, Elevation & Depression
- Line of Sight: The line drawn from the eye of an observer to the point viewed.
- Angle of Elevation: The angle formed by the line of sight with the horizontal when viewing an object above the horizontal level (looking up).
- Angle of Depression: The angle formed by the line of sight with the horizontal when viewing an object below the horizontal level (looking down). Alternate interior angles prove that Angle of Depression equals Angle of Elevation from the target!
2. The Tangent Ratio Method
Most heights-and-distances problems solve using $\tan\theta = \frac{\text{Height}}{\text{Distance}}$:
• If $\theta = 45^\circ \implies \text{Height} = \text{Distance}$ (since $\tan 45^\circ = 1$).
• If $\theta = 30^\circ \implies \text{Height} = \frac{\text{Distance}}{\sqrt{3}}$.
• If $\theta = 60^\circ \implies \text{Height} = \text{Distance} \times \sqrt{3}$.
Visual Learning & Conceptual Map
Some Applications of Trigonometry Master Matrix
Conceptual framework, core mechanisms, and analytical relationships
Academic Architecture
1. Line of Sight, Elevation & Depression • 2. The Tangent Ratio Method
Chapter Summary & 10 Key Takeaways
Takeaway 1
Line of Sight: Direct line from eye to target.
Takeaway 2
Angle of Elevation: Formed when looking up above the horizontal.
Takeaway 3
Angle of Depression: Formed when looking down; equals elevation from the object.
Takeaway 4
Primary Ratio: $\tan\theta = \frac{\text{Opposite (Height)}}{\text{Adjacent (Distance)}}$.
Takeaway 5
Diagram Accuracy: Drawing the correct right triangle is 50% of the solution.
Check Your Understanding (Diagnostic Practice Questions)
Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.
1
A tower stands vertically on the ground. From a point 15 m away from its foot, the angle of elevation of its top is $60^\circ$. Find the height of the tower.
Reveal Answer & Explanation
Answer: $\tan 60^\circ = \frac{h}{15} \implies \sqrt{3} = \frac{h}{15} \implies h = 15\sqrt{3}\text{ meters} \approx 25.98\text{ m}$.
Height = 15*sqrt(3) m.
2
A tree breaks due to a storm and the broken part bends so that the top touches the ground making an angle of $30^\circ$ with it. The distance from foot to point where top touches is 8 m. Find total height of tree.
Reveal Answer & Explanation
Answer: Let unbroken height be $h$, broken fallen part be $l$. $\tan 30^\circ = \frac{h}{8} \implies h = \frac{8}{\sqrt{3}}$. $\cos 30^\circ = \frac{8}{l} \implies \frac{\sqrt{3}}{2} = \frac{8}{l} \implies l = \frac{16}{\sqrt{3}}$. $\text{Total Height} = h + l = \frac{24}{\sqrt{3}} = 8\sqrt{3}\text{ meters}$.
Total height = 8*sqrt(3) m.
3
An observer $1.5\text{ m}$ tall is $28.5\text{ m}$ away from a chimney. The angle of elevation of the top from her eyes is $45^\circ$. Find chimney height.
Reveal Answer & Explanation
Answer: $\tan 45^\circ = \frac{h - 1.5}{28.5} \implies 1 = \frac{h - 1.5}{28.5} \implies h - 1.5 = 28.5 \implies h = 30\text{ meters}$.
Height = 30 m.
4
From the top of a 75 m high lighthouse, the angles of depression of two ships are $30^\circ$ and $45^\circ$. If one ship is directly behind the other, find distance between them.
Reveal Answer & Explanation
Answer: Distance of first ship $= 75 / \tan 45^\circ = 75\text{ m}$. Distance of second ship $= 75 / \tan 30^\circ = 75\sqrt{3}\text{ m}$. Distance between them $= 75(\sqrt{3} - 1)\text{ meters} \approx 54.9\text{ m}$.
Distance = 75(sqrt(3)-1) m.
5
Why is the angle of depression measured from the horizontal line and NOT the vertical wall?
Reveal Answer & Explanation
Answer: Because by scientific definition, all angles of elevation and depression are measured relative to an imaginary horizontal line of sight.
Always measured from horizontal line.
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