| --- |
| source: ncert |
| topic: trigonometry |
| class_level: class_10 |
| chapter: ch9 |
| difficulty: intermediate |
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| |
| Class 10 | Ch9: Applications of Trigonometry (Heights and Distances) |
| KEY TERMS: |
| Angle of elevation: angle formed above horizontal when looking UP at an object. |
| Angle of depression: angle formed below horizontal when looking DOWN at an object. |
| Observer, object, horizontal line form a right triangle. |
| Line of sight: line from eye to object. |
| Note: Angle of elevation from A to B = Angle of depression from B to A (alternate interior angles). |
| STANDARD SETUP: |
| Draw a clear diagram. Identify right triangles. Assign variables. |
| Use appropriate trig ratio: tan for height/distance, sin/cos if hypotenuse involved. |
| STANDARD PROBLEM TYPES: |
| Type 1: Height of tower/building given angle of elevation and distance. |
| Type 2: Distance of boat/object given angle of depression from top. |
| Type 3: Two angles from same point (form two equations). |
| Type 4: Two observers at different points (system of equations). |
| USEFUL: tan(30°)=1/sqrt3, tan(45°)=1, tan(60°)=sqrt3. |
| SOLVED EXAMPLES: |
| Example 1: Tower 75m high. From base, angles of depression of two boats=30° and 45°. |
| Let boat A (30°) distance=x, boat B (45°) distance=y. |
| tan(30°)=75/x → x=75sqrt(3). tan(45°)=75/y → y=75. |
| Distance between boats=x-y=75sqrt(3)-75=75(sqrt3-1)=75(1.732-1)=54.9m. |
| Example 2: Boy 1.2m tall stands 30m from tower. Elevation angle=60°. Find tower height. |
| tan(60°)=(h-1.2)/30 → sqrt3=h-1.2)/30 → h=30sqrt3+1.2=51.96+1.2=53.16m. |
| Example 3: Kite string 150m, angle=60°. Find height of kite. |
| sin(60°)=h/150 → h=150×sqrt3/2=75sqrt3 metres. |
| COMMON MISTAKES: |
| Angle of depression = angle of elevation (they are alternate interior angles, not supplementary). |
| Height of observer must be considered in many problems. |
| Always draw diagram first — most errors come from not visualising correctly. |
