Heights & Distances
Trigonometry steps outside — measuring towers, cliffs and ships you can see but cannot reach, using only angles
- The precise meaning of the angle of elevation and the angle of depression.
- How a single right triangle turns one angle and one length into an unknown height or distance.
- The two-observation method and the standard formula \(h=\dfrac{d}{\cot\alpha-\cot\beta}\).
- Handling an object between two stations, on opposite sides.
- Using the sine and cosine rules when the triangle is not right-angled.
- Reading and using compass bearings in navigation problems.
Angles of Elevation & Depression
When you look at an object, the line of sight is the straight line from your eye to it. The angle of elevation is the angle this line makes above the horizontal; the angle of depression is the angle it makes below the horizontal.
The Right-Triangle Toolkit
Almost every problem reduces to a right triangle whose vertical leg is a height, horizontal leg a distance, and acute angle an elevation or depression. Three ratios do all the work.
For a tower of height \(h\) seen at elevation \(\theta\) from distance \(d\), the master relation is simply \(h=d\tan\theta\), equivalently \(d=h\cot\theta\).
If the observation is made from eye level or from the top of a building, the measured triangle sits above the ground. Add the observer's height back at the end, and always draw the horizontal through the eye, not through the feet.
A Single Observation
The simplest case: one known distance, one measured angle, one unknown. Identify the right triangle, pick the ratio that links what you know to what you want, and solve.
A tower seen at elevation \(30^{\circ}\) from a point \(30\text{ m}\) away has height \(h=30\tan30^{\circ}=10\sqrt3\text{ m}\). One ratio, one line of working.
Two Points on a Straight Line
When the height and the distance are both unknown, a second observation supplies the extra equation. The observer walks a measured distance \(d\) straight toward the tower; the elevation grows from \(\alpha\) to \(\beta\).
Writing each horizontal distance as \(h\cot\alpha\) and \(h\cot\beta\), their difference is the walked distance \(d\). Solving for \(h\) gives the formula — the workhorse of this chapter.
An Object Between Two Stations
If the two observation points lie on opposite sides of the tower, the two horizontal distances add rather than subtract.
When the Triangle Is Oblique
Not every configuration yields a right angle. When the relevant triangle is oblique, fall back on Chapter 13: the sine rule when a side faces a known angle, the cosine rule when an angle is enclosed by two known sides.
Typical triggers: two lines of sight to the same point from different stations (sine rule), or two known legs of a journey with the turn-angle between them (cosine rule for the closing distance).
Bearings & Compass Directions
Navigation problems describe direction by bearing. A compass bearing such as \(N\,30^{\circ}E\) means "start facing North, then turn \(30^{\circ}\) toward East." Three-figure bearings instead measure clockwise from North, from \(000^{\circ}\) to \(360^{\circ}\).
Standard Results
Two configurations recur so often they are worth knowing as formulae, not rederiving each time.
Elevation \(\alpha\) at the first point becomes \(\beta\) after walking \(d\) toward the tower: height \(h=\dfrac{d\tan\alpha\tan\beta}{\tan\beta-\tan\alpha}\).
A staff of height \(a\) on a tower of height \(h\) subtends angle \(\beta\) at a point where the tower subtends \(\alpha\): \(h=\dfrac{a\tan\alpha}{\tan(\alpha+\beta)-\tan\alpha}\).
If the shadow lengthens by \(d\) as the sun's elevation drops from \(\beta\) to \(\alpha\), then \(h=\dfrac{d}{\cot\alpha-\cot\beta}\) — the same workhorse formula.
Putting It to Work
Problem. The elevation of the top of a tower from a point \(30\text{ m}\) from its foot is \(30^{\circ}\). Find the height.
Solution. With \(h=d\tan\theta\):
Problem. From the top of a \(100\text{ m}\) cliff the angles of depression of two boats in line with the foot are \(30^{\circ}\) and \(45^{\circ}\). Find the distance between them.
Solution. Distances from the foot are \(100\cot30^{\circ}\) and \(100\cot45^{\circ}\):
Problem. The elevation of a tower is \(30^{\circ}\) at a point; on walking \(40\text{ m}\) closer it becomes \(60^{\circ}\). Find the height.
Solution. Use \(h=\dfrac{d\tan\alpha\tan\beta}{\tan\beta-\tan\alpha}\) with \(\alpha=30^{\circ},\ \beta=60^{\circ},\ d=40\):
Problem. A tower stands between two points \(100\text{ m}\) apart on opposite sides; the elevations are \(30^{\circ}\) and \(60^{\circ}\). Find the height.
Solution. Use the opposite-side formula \(h=\dfrac{d}{\cot\alpha+\cot\beta}\):
Problem. From the foot of a tower the top of a hill is seen at elevation \(60^{\circ}\); from the foot of the hill the top of the tower is seen at \(30^{\circ}\). If the tower is \(50\text{ m}\) high, find the height of the hill.
Solution. Let the horizontal separation be \(d\). From the hill's foot, \(\tan30^{\circ}=\dfrac{50}{d}\Rightarrow d=50\sqrt3\); from the tower's foot, hill height \(H=d\tan60^{\circ}\):
Problem. A ship sails \(12\text{ km}\) due north, then \(5\text{ km}\) due east. Find its distance and bearing from the start.
Solution. The two legs are perpendicular, so the closing distance is the hypotenuse:
Chapter Summary
Elevation above horizontal, depression below; depression from above \(=\) elevation from below.
\(h=d\tan\theta\); add the observer's height where needed.
\(h=\dfrac{d}{\cot\alpha-\cot\beta}\) for a walk toward the object.
\(h=\dfrac{d}{\cot\alpha+\cot\beta}\) when the object is between the stations.
Sine rule for a side facing a known angle; cosine rule for an enclosed angle.
Resolve each leg into North–South and East–West components first.
Problems
Draw the figure first, label the right triangle, then choose the ratio or rule. Difficulty rises down the list.
- The elevation of the top of a pole from a point \(20\text{ m}\) away is \(45^{\circ}\). Find its height.
- A ladder \(10\text{ m}\) long leans against a wall at \(60^{\circ}\) to the ground. How high up the wall does it reach?
- From a \(60\text{ m}\) cliff the depression of a boat is \(30^{\circ}\). How far is the boat from the foot of the cliff?
- The elevation of a tower from a point is \(30^{\circ}\); from a point \(50\text{ m}\) nearer it is \(60^{\circ}\). Find the height.
- A tree breaks; its top touches the ground \(10\text{ m}\) from the base, making \(30^{\circ}\) with the ground. Find the original height.
- From the top of a \(15\text{ m}\) building the elevation of a tower's top is \(30^{\circ}\) and the depression of its foot is \(60^{\circ}\). Find the tower's height.
- The shadow of a tower is \(30\text{ m}\) longer at elevation \(30^{\circ}\) than at \(60^{\circ}\). Find the height.
- A ship sails \(8\text{ km}\) west, then \(6\text{ km}\) north. Find its distance and bearing from the start.
- Two pillars of equal height stand on either side of a \(100\text{ m}\) road; from a point between them the elevations are \(30^{\circ}\) and \(60^{\circ}\). Find the height and the position of the point.
- The elevation of a tower at a point \(A\) due south is \(45^{\circ}\) and at \(B\) due west of \(A\) is \(30^{\circ}\). If \(AB=60\text{ m}\), find the height.
- From a point the elevation of a tower's top is \(\alpha\), and a flagstaff of height \(a\) on it subtends \(\beta\). Show the tower's height is \(\dfrac{a\tan\alpha}{\tan(\alpha+\beta)-\tan\alpha}\).
- An aeroplane at height \(h\) observes two points on a line below at depressions \(\alpha\) and \(\beta\). Find the distance between the points.