source: ncert
topic: trigonometry
class_level: class_10
chapter: ch9
difficulty: intermediate
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.