Hello, and welcome to today's mathematics lesson. We are going to explore heights and distances. This chapter brings together your knowledge of trigonometry and applies it to real-world situations — measuring things that are too tall to climb, or too far away to reach. By the end of this lesson, you will understand how to use angles of elevation and depression to calculate unknown heights and distances.
Let us begin with the fundamental concepts. Imagine you are standing on level ground, looking up at the top of a tower. The straight line from your eye to the top of the tower is called the line of sight. Now, the line of sight forms an angle with the horizontal line drawn through your eye level. This angle is called the angle of elevation.
Formally, the angle of elevation is defined as the angle which the line of sight makes with the horizontal when the observer is looking upwards at an object.
Conversely, imagine you are at the top of that same tower, looking down at a point on the ground. The angle that your line of sight makes with the horizontal, when you are looking downwards, is called the angle of depression.
Here is a crucial relationship you must remember. The angle of elevation of a point as seen from another point is equal to the angle of depression of the second point as seen from the first. This is because they are alternate angles formed by parallel horizontal lines. So if the angle of elevation of the top of a tower from a point on the ground is 30 degrees, then the angle of depression of that ground point from the top of the tower is also 30 degrees.
Now, let us see how we apply these concepts. The key is to form right-angled triangles and use trigonometric ratios. Remember your three fundamental ratios. Sine of an angle equals opposite over hypotenuse, sin θ = opp/hyp. Cosine of an angle equals adjacent over hypotenuse, cos θ = adj/hyp. And tangent of an angle equals opposite over adjacent, tan θ = opp/adj.
In height and distance problems, we most frequently use the tangent ratio, because we typically know — or want to find — the horizontal distance and the vertical height. So the basic working formula becomes: tan θ = h/d.
Let me walk you through a classic example. Suppose the length of the shadow of a vertical tower is √3 times its height. We need to find the angle of elevation of the sun.
Let the height of the tower be x. Then the length of the shadow is √3·x metres. If theta is the angle of elevation of the sun, then tan θ = x/(√3·x) = 1/√3. We know that tan 30° = 1/√3. Therefore, θ = 30°.
Here is another situation you will often encounter. Two observers stand on the same side of a tower, in a straight line with it. They measure angles of elevation of 25 degrees and 50 degrees to the top. How do we find the distance between them?
Let us say the tower height is 70 m. From the first observer, tan 25° = 70/AQ, so AQ = 70/tan 25° ≈ 150.118 m. From the second observer, tan 50° = 70/BQ, so BQ = 70/tan 50° ≈ 58.735 m. The distance between the two people is AQ − BQ, which equals approximately 91.38 m.
Problems involving the angle of depression require careful interpretation. Remember that the angle of depression is measured from the horizontal looking down. But when we draw our right-angled triangle, we use the fact that this angle equals the angle of elevation from the object below.
Consider an observation tower 180 metres above sea level. An enemy boat is observed at an angle of depression of 29 degrees. To find the distance of the boat from the foot of the tower, we use the fact that the angle of elevation from the boat to the top of the tower is also 29 degrees. So tan 29° = 180/PB, giving us PB = 180/tan 29° ≈ 325 m.
Some problems involve changing conditions — a moving object or a changing angle of the sun. These require setting up equations and solving them systematically.
For instance, consider a tower whose shadow changes as the sun moves. When the sun's altitude is 45 degrees, the shadow is shorter. When the sun's altitude drops to 30 degrees, the shadow becomes longer by 10 metres. We need to find the height of the tower.
Let the height be h. At 45 degrees, tan 45° = h/BC, so BC = h. At 30 degrees, tan 30° = h/BD, so BD = h√3. The difference BD − BC = 10, so h√3 − h = 10. Factoring: h(√3 − 1) = 10. Rationalising: h = 10(√3+1)/[(√3−1)(√3+1)] = 10(√3+1)/2 = 5(√3+1) ≈ 13.66 m.
Another elegant problem involves two ships observed from a cliff. From the top of a 200 metre cliff, the angles of depression to two ships are 45 degrees and 30 degrees.
If the ships are on the same side of the cliff, the distance between them is the difference of their distances from the foot. For the nearer ship at 45 degrees: tan 45° = 200/BC, so BC = 200 m. For the farther ship at 30 degrees: tan 30° = 200/BD, so BD = 200√3 ≈ 346.4 m. The distance between ships is 346.4 − 200 = 146.4 m.
If the ships are on opposite sides, we add the distances: 346.4 + 200 = 546.4 m.
Problems involving speed and time require relating distance, speed, and time with your trigonometric equations.
Consider a car approaching a tower. It takes 12 minutes for the angle of depression to change from 30 degrees to 45 degrees. How much more time until it reaches the tower?
Let the speed be x m/min and the remaining time be t min. From the geometry: h = tx from the 45 degree position, and h/(12x + tx) = tan 30° = 1/√3 from the initial position. Substituting and simplifying: √3 t = 12 + t, so t = 12/(√3−1) = 12(√3+1)/2 = 6(√3+1) ≈ 16.39 min.
A particularly beautiful problem involves a cloud and its reflection in a lake. From a point 25 metres above the lake, the angle of elevation of a cloud is 30 degrees, while the angle of depression of its reflection is 60 degrees.
Let the height of the cloud above the lake be h. The cloud is h − 25 metres above the observer. The reflection is h + 25 metres below the observer's horizontal level. From the geometry, the horizontal distance PE can be expressed two ways: PE = √3(h−25) from the elevation triangle, and PE = (h+25)/√3 from the depression triangle. Equating: 3(h−25) = h+25, yielding 3h − 75 = h + 25, so h = 50 m.
Finally, let us consider problems with two angles given as tangents. From a point on the ground, the tangent of the angle of elevation of a tower is 3/5. After walking 50 metres towards the tower, the tangent of the new angle becomes 4/5.
Let the height be h and the final distance be x. Then 3/5 = h/(x+50) and 4/5 = h/x. From the second equation: h = 4x/5. Substituting into the first: 3/5 = (4x/5)/(x+50), which simplifies to 3(x+50) = 4x, giving 3x + 150 = 4x, so x = 150 m. Therefore h = 4×150/5 = 120 m.
Let me now summarise the key takeaways from this chapter.
First, the angle of elevation is measured upward from the horizontal, while the angle of depression is measured downward from the horizontal. These two angles are equal when observing the same two points from opposite positions.
Second, always identify or construct right-angled triangles in your diagram. The tangent ratio is most commonly used since problems typically involve vertical heights and horizontal distances.
Third, for problems with changing conditions, define variables clearly and set up equations based on the geometry at each stage.
Fourth, when dealing with reflections in water, remember that the image appears as far below the surface as the object is above it.
Fifth, for moving objects, relate distance, speed, and time through your trigonometric relationships.
And finally, always check whether your answer is reasonable — towers should be tens or hundreds of metres, not thousands, unless specified otherwise.
Heights and distances is one of the most practical applications of trigonometry you will study. Master these techniques, and you can estimate the height of a tree, the width of a river, or the distance to a ship at sea — all without leaving the ground. Keep practising, stay curious, and I look forward to seeing you in the next lesson.