Introduction
This chapter applies trigonometric ratios to practical problems involving heights and distances. You will learn to find the height of a tower, the width of a river, or the distance of a ship from a lighthouse using angles of elevation and depression. These problems are based on right-angled triangles formed between the observer, the object, and the ground.
Angle of Elevation and Depression
The angle of elevation is the angle formed between the horizontal line and the line of sight when an observer looks UP at an object. The angle of depression is the angle formed between the horizontal line and the line of sight when an observer looks DOWN at an object. Note that the angle of elevation from one point equals the angle of depression from the other point (alternate interior angles with the horizontal).
Key Points
- •Angle of elevation: observer looks UP from horizontal
- •Angle of depression: observer looks DOWN from horizontal
- •Both are measured from the horizontal, never from the vertical
- •Angle of elevation from A = Angle of depression from B (alternate angles)
- •Always draw a clear diagram before solving
Watch Out
The most common mistake is measuring the angle from the vertical instead of the horizontal. Always draw the horizontal line first.
Problems on Heights and Distances
These problems involve finding unknown heights, distances, or angles using trigonometric ratios in right-angled triangles. The general approach is: (1) Draw a clear diagram, (2) Identify the right triangle, (3) Label all known and unknown quantities, (4) Choose the appropriate trigonometric ratio, (5) Set up the equation and solve. Problems may involve one or two right triangles.
Key Points
- •tan is used most often (relates height and distance on ground)
- •sin and cos are used when hypotenuse (slant distance) is involved
- •In two-triangle problems, use a common side to connect equations
- •When angles from both sides are given, form equations with the same unknown
- •Heights of buildings/towers are always measured from the ground
Worked Example
A tower is 50 m high. The angle of elevation from a point on the ground is 30 degrees. Find the distance of the point from the base. Let the distance = d metres. tan 30 = height/distance = 50/d 1/sqrt(3) = 50/d d = 50.sqrt(3) = 50 x 1.732 = 86.6 m
Watch Out
Use tan for most height-distance problems. Use sin or cos only when the line of sight (hypotenuse) is involved.
Two-Triangle Problems
Some problems involve two right-angled triangles sharing a common side. For example, finding the height of a tower when angles of elevation from two points at different distances are given. In these cases, set up two equations using the two triangles and solve simultaneously. The common side (usually the height) connects the two equations.
Key Points
- •Set up separate equations for each triangle
- •Use the same variable for the common unknown
- •Eliminate the unwanted variable to find the required one
- •These are typically 4 or 5 mark questions in exams
- •The key is a good diagram with all angles and sides labelled
Worked Example
The angles of elevation of the top of a tower from two points at distances d1 and d2 from the base (d1 > d2) are 30 and 60 degrees respectively. Find the height of the tower. Let h = height of tower. From point 1: tan 30 = h/d1, so d1 = h.sqrt(3) From point 2: tan 60 = h/d2, so d2 = h/sqrt(3) If d1 - d2 is given, substitute and solve for h.
Quick Summary
- ✓Angle of elevation: looking UP from horizontal
- ✓Angle of depression: looking DOWN from horizontal
- ✓Always draw a diagram and identify the right triangle
- ✓tan(angle) = height/distance is the most common setup
- ✓For two-triangle problems, use a common side to link equations
- ✓Standard angles (30, 45, 60) appear most frequently
Key Formulas
tan(theta) = Opposite/Adjacent = Height/Distance
sin(theta) = Opposite/Hypotenuse
cos(theta) = Adjacent/Hypotenuse
tan 30 = 1/sqrt(3), tan 45 = 1, tan 60 = sqrt(3)
sqrt(3) approx 1.732
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