3.3 Applications of trigonometry

Most application problems are given in words, and the diagram is the most important step of the solution. A good diagram shows every given length and angle in the right place, marks every right angle, labels the unknown, and is roughly to scale, so that an answer that is obviously wrong stands out.

A reliable routine for drawing one:

1. Draw the horizontal ground (or sea level) first, then vertical objects such as towers, cliffs and people at right angles to it.

2. Place each point named in the question and label it with its letter.

3. Mark every given length and angle, with units, and mark right angles with a small square.

4. Label the unknown with a letter such as $x$, $h$ or $\theta$.

5. Look for triangles that contain the unknown and enough known information to solve them.

Angles of elevation and depression

When an observer looks up at an object, the line of sight rises above the horizontal; when they look down, it falls below it. The angle is always measured from the horizontal, never from the vertical.

In the figure, an observer at $A$ on top of a cliff looks down at a boat $B$. The angle of depression of $B$ from $A$ is measured below the horizontal through $A$. From the boat, the angle of elevation of $A$ is measured above the sea, which is also horizontal.

The two horizontal lines are parallel, and the line of sight crosses both of them. The angle of depression and the angle of elevation are therefore alternate angles, and they are equal. This is useful because the angle of depression is outside the right-angled triangle formed by the cliff, the sea and the line of sight, while the equal angle of elevation is inside it.

Heights of objects that cannot be reached

The method in the Eiffel Tower example needs the distance to the foot of the object. Often this distance is unknown: the foot of a mountain is inside the mountain, and the base of a tower may be behind a fence. Measuring two angles of elevation from two points a known distance apart, in line with the object, solves the problem.

There are two standard methods.

Method 1: the sine rule

In triangle $ABT$, the angle at $B$ is $180° - \beta$, so the angle at $T$ is $180° - \alpha - (180° - \beta) = \beta - \alpha$. Equivalently, $\beta$ is an exterior angle of triangle $ABT$, so it equals the sum of the two interior opposite angles, $\alpha$ and $A\hat{T}B$. With side $d$ and its opposite angle $\beta - \alpha$ known, the sine rule gives $BT$, and then $h = BT\sin\beta$ in the right-angled triangle $BFT$.

Method 2: two equations

Each right-angled triangle gives an expression for $h$: $h = x\tan\beta$ from triangle $BFT$, and $h = (x + d)\tan\alpha$ from triangle $AFT$. Setting them equal gives one equation in $x$. This method suits non-calculator questions with exact values of $\tan$.

Bearings

A bearing describes a direction on a map or chart. Bearings are used in navigation at sea and in the air, in orienteering and in surveying, because they give a direction without reference to any landmark.

A bearing always has three digits, so a direction $45°$ clockwise from north is written $045°$. The four main compass directions are north $000°$, east $090°$, south $180°$ and west $270°$; north-east is $045°$, south-east $135°$, south-west $225°$ and north-west $315°$.

Reading "the bearing of B from A"

The phrase "the bearing of $B$ from $A$" means: stand at $A$, face north, and turn clockwise until facing $B$. The point after "from" is where the angle is measured. Draw a north line at that point before measuring anything.

Back bearings

The north lines at $A$ and $B$ are parallel. Going from $A$ to $B$ and then back from $B$ to $A$ means facing in exactly the opposite direction, which is a turn of $180°$. So the bearing of $A$ from $B$, the back bearing, is found by adding or subtracting $180°$.

For example, if the bearing of a port from a ship is $305°$, the bearing of the ship from the port is $305° - 180° = 125°$.

Finding angles inside a triangle from bearings

Bearings give angles measured from north lines, while the sine and cosine rules need the angles inside a triangle. To convert, draw a north line at each turning point and use parallel-line facts: co-interior angles between two north lines add to $180°$, and the back bearing at the new point gives the direction back to the previous point. The interior angle at a turning point is the difference between the back bearing and the new bearing (or $360°$ minus it, if that is the angle inside the triangle).

When the turn is not a right angle, the triangle is solved with the cosine rule for the distance and the sine rule for the angle.

Surveying and triangulation

Distances across rivers, valleys and lakes cannot be measured with a tape. Surveyors measure a baseline on accessible ground instead, then measure the angles from each end of it to the distant point.

Triangulation fixes a point with only one measured length, because two angles and the side between them determine a triangle completely. A chain of triangles, each sharing a side with the next, can carry measurements across a whole country. The heights of Himalayan peaks, including Mount Everest, were first calculated in the nineteenth century by triangulation from survey stations on the plains of northern India, more than 150 km away.

The width could also be found from the area of the triangle: the area is $\frac{1}{2} \times AB \times AC \times \sin A$, and also $\frac{1}{2} \times AB \times w$, so $w = AC\sin A$, the same calculation. Many surveying problems have more than one route to the answer, and a second method is a good check.

Problems in three dimensions

Three-dimensional problems combine everything above. The key facts are that a vertical line is perpendicular to every horizontal line through its foot, and that two lines on level ground meeting at a known angle (often given by bearings such as "due west" and "due south") lie in a horizontal plane. Each triangle needed is drawn separately, flat on the page, with its right angle marked.

A strategy for problems in three dimensions:

1. Draw a sketch of the whole situation, with horizontal ground drawn as a parallelogram (as if seen at an angle) and vertical lines drawn upright.

2. Identify each triangle that contains a right angle: usually one vertical side and one horizontal side.

3. Solve the triangles in a chain, carrying unrounded values from one to the next.

4. Use the sine or cosine rule for any triangle without a right angle, such as one lying on the ground or one tilted in space.

When the angle at the foot of a vertical object is not a right angle, the triangle on the ground needs the cosine rule, and an angle between two sloping lines needs it again in a tilted triangle.

Exam-style questions on applications of trigonometry

The next question is set without a calculator and uses the exact values $\tan 30° = \frac{1}{\sqrt{3}}$, $\tan 60° = \sqrt{3}$ and $\sin 60° = \frac{\sqrt{3}}{2}$ (derived in 3.5).

Self-check

• To do:

  I can draw a clear, labelled diagram from a written description, with right angles marked and the unknown labelled.

• To do:

  I can use angles of elevation and depression, including the fact that they are equal as alternate angles.

• To do:

  I can allow for the height of an observer's eye above the ground.

• To do:

  I can find the height of an object whose foot cannot be reached, from two angles of elevation.

• To do:

  I can write bearings with three figures, find back bearings, and find the angle inside a triangle at a turning point.

• To do:

  I can solve navigation problems with Pythagoras's theorem, the sine rule and the cosine rule, and give the final direction as a bearing.

• To do:

  I can find an inaccessible distance by triangulation.

• To do:

  I can solve problems in three dimensions by finding the right-angled triangles and solving them in a chain.

1. (a)

   Find $\mathrm{AC}$.

   [3]3 marks

   Markscheme (a)

   attempt to use the cosine rule (M1)

   $\mathrm{AC}^2 = 85^2 + 110^2 - 2(85)(110)\cos 104°$ (A1)

   $\mathrm{AC} = 154$ (m) $(154.431\ldots)$ A1

2. (b)

   Find $\mathrm{B\hat{A}C}$.

   [3]3 marks

   Markscheme (b)

   $\frac{\sin\mathrm{B\hat{A}C}}{110} = \frac{\sin 104°}{154.431\ldots}$

3. (c)

   It is also known that $\mathrm{CD} = 70$ m and $\mathrm{A\hat{C}D} = 57°$. Find $\mathrm{AD}$.

   [3]3 marks
4. (d)

   Find the area of the field.

   [4]4 marks

   Markscheme (d)

   area of triangle $\mathrm{ABC} = \frac{1}{2}(85)(110)\sin 104° = 4536.03\ldots$

5. (e)

   The farmer puts a fence around the boundary of the field. The fence costs $14.50 per metre. Find the cost of the fence, correct to the nearest dollar.

   [2]2 marks

   Markscheme (e)

   perimeter $= 85 + 110 + 70 + 130.282\ldots = 395.282\ldots$ (M1)

   cost $= \$5732$

(b) The new distance is $422.38\ldots - 100 = 322.38\ldots$ m.

Moving closer makes the angle of elevation larger, as expected.

In the right-angled triangle $BFT$, $h$ is opposite the $41°$ angle and $BT$ is the hypotenuse.

Method 2. Let $BF = x$. Then $h = x\tan 41° = (x + 50)\tan 28°$.

Both methods give the same height.

(a) Use Pythagoras's theorem.

(b) In the right-angled triangle, the angle at $A$ satisfies

$C$ is clockwise from the direction $AB$, so add this angle to the bearing of $B$.

(a) Two sides and the included angle are known: cosine rule.

(b) Find angle $BAC$ with the sine rule. It is opposite $BC = 15$, which is not the longest side, so it is acute.

The ship turned clockwise at $B$, so $C$ is clockwise from the direction of $B$ as seen from $A$.

(c) The return journey is the back bearing.

It is opposite the baseline $AB = 200$ m, so the sine rule can be used. $AC$ is opposite the $71°$ angle.

The width $w$ is the perpendicular distance $CD$. In the right-angled triangle $ADC$, $AC$ is the hypotenuse and $w$ is opposite the $62°$ angle.

(b) In the vertical triangle $BFT$, also with a right angle at $F$:

$B$ is further from the mast than $A$, so its angle of elevation is smaller.

(c) West and south are perpendicular, so triangle $AFB$ on the ground has a right angle at $F$.

From $A$, the point $B$ is $60$ m east and $80$ m south. The angle between south and $AB$ is $\tan^{-1}\left(\frac{60}{80}\right) = 36.869\ldots°$, measured from south towards east, which is anticlockwise from south.

(b) Triangle $AFB$ lies on the ground with the included angle $75°$ at $F$: cosine rule.

(c) Triangle $ATB$ is tilted in space and has no right angle, but all three sides are now known: cosine rule.

Using $TB^2 = 10\,600$ and $AB^2$ unrounded keeps the answer accurate.

(b) Two sides and the included angle: cosine rule.

(c) First find angle $QPR$ with the sine rule ($QR = 11$ is opposite it; $11$ is not the longest side, so the angle is acute).

The bearing of $R$ from $P$ is $050° + 60.656\ldots° = 110.656\ldots°$. The bearing of $P$ from $R$ is the back bearing.

(d) The route via $Q$ is $8 + 11 = 19$ km, and the direct route is $12.427\ldots$ km. The time saved is

Triangle $ABT$ has two angles of $30°$ (at $A$ and at $T$), so it is isosceles, with $BT = AB = 40$ m (the sides opposite the equal angles).

(b) In the right-angled triangle $BFT$, $BT = 40$ m is the hypotenuse and the height $FT$ is opposite the $60°$ angle.

which is the given result. As a check with Method 2: $FT = BF\tan 60° = (BF + 40)\tan 30°$ gives $\sqrt{3}\,BF = \frac{BF + 40}{\sqrt{3}}$, so $3BF = BF + 40$, $BF = 20$ and $FT = 20\sqrt{3}$.

$\mathrm{B\hat{A}C} = 43.7°$ $(43.7199\ldots)$ A1

Markscheme (c)

area of triangle $\mathrm{ACD} = \frac{1}{2}(154.431\ldots)(70)\sin 57° = 4533.18\ldots$ (A1)

total area $= 9070$ m$^2$ $(9069.21\ldots)$ A1

Note: Award (M1) for a correct method for either triangle.

Note: Award (M1)A0 for $5731.60 (not to the nearest dollar) or $5728 (from $\mathrm{AD} = 130$).

[2]2 marks