Class 10 NCERT Maths 9: Applications of Trigonometry
This chapter focuses on solving real-life problems using the trigonometric ratios you learned in Chapter 8. We mainly deal with heights and distances in situations involving right-angled triangles.
Key Terms
- Line of Sight: The imaginary line from your eye to an object you are looking at.
- Angle of Elevation: Angle formed when you look upward at an object from a horizontal line.
- Angle of Depression: Angle formed when you look downward at an object from a horizontal line.
Which Ratio to Use?
Use the right-angled triangle made by the observer, the object, and the horizontal ground.
| Given | Required | Use | Formula |
|---|---|---|---|
| Perpendicular (p) and Hypotenuse (h) | Find angle or missing side | Sine | \(\sin\theta = p/h\) |
| Base (b) and Hypotenuse (h) | Find angle or missing side | Cosine | \(\cos\theta = b/h\) |
| Perpendicular (p) and Base (b) | Find angle or missing side | Tangent | \(\tan\theta = p/b\) |
Common Approach to Solve Problems
- Draw a diagram based on the problem statement.
- Identify p, b, and h.
- Choose the correct trigonometric ratio (\(p/h\), \(b/h\), \(p/b\), etc.).
- Substitute the given values and solve.
Example Problem
Example: A man standing on the ground observes the top of a tower at an angle of elevation of \(30^\circ\). If the tower is 20 m high, find the distance from the man to the base of the tower.
Here: Perpendicular \(p = 20 \text{ m}\) Angle \(\theta = 30^\circ\) Base \(b = ?\) We use: \[ \tan\theta = \frac{p}{b} \] \[ \tan 30^\circ = \frac{20}{b} \] \[ \frac{1}{\sqrt{3}} = \frac{20}{b} \] \[ b = 20\sqrt{3} \ \text{m} \]
Quick Tips
- Always keep \(p\), \(b\), \(h\) consistent with your triangle diagram.
- Angles of elevation and depression are measured from the horizontal.
- Use \(\sin\theta = p/h\), \(\cos\theta = b/h\), \(\tan\theta = p/b\) for quick recall.