Pythagoras and trigonometry. Attnom Academy

M4.3.1 Chapter 1

Pythagoras' theorem

A triangle has a right angle and two short sides of 3 cm and 4 cm. How long is the third side?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A 7 cm, because 3 + 4 = 7.
You picked A. Read on. At the end, the chapter comes back to your guess.
B 5 cm, because 3 × 3 + 4 × 4 = 5 × 5.
You picked B. Read on. At the end, the chapter comes back to your guess.
C 6 cm, because the third side is between 3 and 4, and then a bit more.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Find the longest side of a right angled triangle
Find a shorter side when you know the other two
Show that sides 5, 12 and 13 make a right angle

A right angled triangle has one angle of 90°. The side opposite that angle is the longest side. It is called the hypotenuse. The two shorter sides, which meet at the right angle, are the legs.

The hypotenuse faces the right angle. It is always the longest side.

Draw a square on each side of a right angled triangle. Then compare the areas. Ana does this with a triangle whose legs are 4 cm and 3 cm.

Draw it first

Squares on the sides

1 Draw a right angled triangle. Its legs are 4 cm and 3 cm. Draw a square on the 4 cm leg. Its area is 4 × 4 = 16 cm².
2 Draw a square on the 3 cm leg. Its area is 3 × 3 = 9 cm².
3 Draw a square on the hypotenuse. Measure the hypotenuse: it is 5 cm. The square has area 5 × 5 = 25 cm².

Look at the three areas: 16 + 9 = 25. The two small squares have the same total area as the big square. This is true for every right angled triangle. It is called Pythagoras' theorem.

Why is it true? Draw two squares with the same side, a + b. Each holds four copies of the triangle, all with legs a and b. Take the four triangles away from each square.

Same big square, same four triangles. So the white parts have equal area: a² + b² = c².

The big squares are equal. The four triangles are equal. So what is left must be equal too: a² + b² on the left, c² on the right. That is a proof, not a guess.

In a right angled triangle, a² + b² = c². Here c is the hypotenuse. The rule works only when the triangle has a right angle.

To find the hypotenuse, add the squares of the legs, then take the square root. We take the positive root, because a length is above 0.

Worked example

A right angled triangle has legs of 6 cm and 8 cm. Find the hypotenuse.

The legs are a = 6 and b = 8. We want the hypotenuse c.
Use c² = a² + b² = 6² + 8².
Square each leg and add: 36 + 64 = 100.
Take the square root: c = √100 = 10.

Answer: The hypotenuse is 10 cm.

The answer is not always a whole number. When it is not, the calculator gives a long decimal. Say how you rounded.

Worked example

A right angled triangle has legs of 5 cm and 7 cm. Find the hypotenuse to 1 decimal place.

Use c² = 5² + 7².
Square each leg: 5² = 25 and 7² = 49.
Add: c² = 25 + 49 = 74.
Take the square root: c = √74 = 8.602 on the calculator. To 1 decimal place that is 8.6.

Answer: To 1 decimal place, the hypotenuse is 8.6 cm.

Now the hypotenuse is known and a leg is not. Rewrite the rule. If c² = a² + b², then a² = c² − b². So you subtract. The hypotenuse is always the one you start from.

Worked example

A right angled triangle has hypotenuse 25 cm and one leg 7 cm. Find the other leg.

The hypotenuse is c = 25. One leg is b = 7. We want the other leg a.
Use a² = c² − b² = 25² − 7².
Square each number: 25² = 625 and 7² = 49.
Subtract: a² = 625 − 49 = 576.
Take the square root: a = √576 = 24.

Answer: The other leg is 24 cm.

The rule also works backwards. Suppose a triangle has sides 5 cm, 12 cm and 13 cm. Is there a right angle? Try the sum with c as the longest side.

Is the corner between the 5 cm and 12 cm sides a right angle? The sums will tell us.

Worked example

Do sides of 5 cm, 12 cm and 13 cm make a right angled triangle?

The longest side is 13. If there is a right angle, it is opposite this side, so c = 13.
The squares of the two shorter sides are 5² = 25 and 12² = 144.
Add them: 25 + 144 = 169.
The square of the longest side is 13² = 169.
The two results are equal, so a² + b² = c² holds. The triangle has a right angle.

Answer: Yes. 25 + 144 = 169, and 13 × 13 = 169, so the triangle has a right angle.

The sums matched, so the corner opposite the 13 cm side is a right angle. The longest side faces it.

Why can we trust it? A triangle is fixed by its three sides. Build the right angled triangle with legs 5 and 12. Its hypotenuse is 13. Any triangle with sides 5, 12 and 13 is the same triangle, so it has a right angle too. The sums must match exactly. Sides 4, 5 and 6 give 16 + 25 = 41, but 6 × 6 = 36. So they make no right angle.

See it three ways

With things

A builder ties a loop of string with 12 equal spaces. Pulled tight with 3 spaces, 4 spaces and 5 spaces, it makes a square corner.

As a picture

In symbols

3² + 4² = 9 + 16 = 25 = 5².

Try it

A square corner from string

You need: a piece of string about 60 cm long, a pencil and a ruler, a book with a square corner.

Mark the string every 5 cm, so there are 12 equal spaces. Tie the ends together.
Hold the loop at three marks so the sides are 3 spaces, 4 spaces and 5 spaces long. Pull it tight.
Put the corner of the book into the corner between the 3 and the 4. Compare.

What you will see: The corner of the string fits the corner of the book. 3, 4 and 5 make a right angle, because 3² + 4² = 5².

If you cannot try it, this is what happens.

A wrong idea many people have

a² + b² = c² works for every triangle.

Why it feels right: The first triangles we meet in this chapter all have right angles. So the rule seems to belong to every triangle.

The test that breaks it: Take a triangle with sides 4, 5 and 7. Then 4² + 5² = 16 + 25 = 41, but 7² = 49. The two results are not equal.

What is true: The rule holds only when the triangle has a right angle, and c is the side opposite it. When the sums do not match, there is no right angle.

A wrong idea many people have

The hypotenuse is a + b, because it joins the ends of the two legs.

Why it feels right: Walking along both legs and walking across the corner look like the same trip.

The test that breaks it: In the 3, 4, 5 triangle, a + b = 3 + 4 = 7, but the hypotenuse is 5 cm. The short cut across is shorter than the way round.

What is true: The hypotenuse is less than a + b. Square first, add, then take the root: c = √(a² + b²).

Key words

hypotenuse: the longest side of a right angled triangle, opposite the right angle.
Pythagoras' theorem: in a right angled triangle, a² + b² = c², with c the hypotenuse.

Back to your guess

The answer is B: 5 cm, because 3 × 3 + 4 × 4 = 5 × 5.

The third side is 5 cm. It cannot be 7 cm: in a triangle, one side is always shorter than the other two added together. The sum 3 × 3 + 4 × 4 = 25 = 5 × 5 is Pythagoras' theorem, and this chapter shows why it is true.

Keep it

Copy this into your exercise book.

In a right angled triangle, a² + b² = c². The hypotenuse c faces the right angle. Add squares to find c. Subtract squares to find a leg.

Draw: A right angled triangle with a square on each side, labelled a², b² and c², and a² + b² = c² under it.

M4.3.2 Chapter 2

Using Pythagoras

A box is 2 cm by 3 cm by 6 cm. How long is the longest stick that fits inside it?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A 6 cm, because no stick can be longer than the longest edge.
You picked A. Read on. At the end, the chapter comes back to your guess.
B 11 cm, because 2 + 3 + 6 = 11 and the stick can bend round.
You picked B. Read on. At the end, the chapter comes back to your guess.
C 7 cm, from one bottom corner to the opposite top corner.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Find how high a ladder reaches up a wall
Find the distance between two points on a grid
Find the longest stick that fits inside a box

A wall stands straight up from flat ground. So the wall and the ground meet at a right angle. A ladder leaning on the wall makes a right angled triangle with them. The ladder is the hypotenuse.

Draw it first

Find the triangle in the ladder

1 A ladder leans on a wall. Its foot is 0.9 m from the wall. Its top reaches a height h.
2 Draw the wall, the ground and the ladder as a triangle. The ladder, 4.1 m, is the hypotenuse.

You know the hypotenuse and one leg, so you subtract the squares. The height is h² = 4.1² − 0.9².

Worked example

A ladder 4.1 m long stands with its foot 0.9 m from a wall. How high up the wall does it reach?

The ladder is the hypotenuse: c = 4.1. The foot is b = 0.9 from the wall. We want the height h.
Use h² = c² − b² = 4.1² − 0.9².
Square each number: 4.1² = 16.81 and 0.9² = 0.81.
Subtract: h² = 16.81 − 0.81 = 16.
Take the square root: h = √16 = 4.

Answer: The ladder reaches 4 m up the wall.

A grid hides a right angled triangle too. To find how far apart two points are, go across, then go up. The two moves are the legs. The straight line between the points is the hypotenuse.

A is at (1, 2) and B is at (7, 10). One square is 1 km.

Worked example

A walker goes from A (1, 2) to B (7, 10) on a map. One square is 1 km. How far apart are A and B in a straight line?

Across from A to B: 7 − 1 = 6 km east.
Up from A to B: 10 − 2 = 8 km north.
The two moves are the legs of a right angled triangle. The distance d is its hypotenuse.
d² = 6² + 8² = 36 + 64 = 100.
d = √100 = 10 km.

Answer: The distance is 10 km.

Take care when a coordinate is negative. To go from −2 to 3 you move 3 − (−2) = 5 places. You always take the smaller number away from the larger one.

P is at (−2, −3) and Q is at (3, 2). One square is 1 km, and x and y are the kilometres east and north.

Worked example

Two huts are at P (−2, −3) and Q (3, 2) on a map. One square is 1 km. How far apart are they, to 1 decimal place?

Across from P to Q: 3 − (−2) = 5 km.
Up from P to Q: 2 − (−3) = 5 km.
The two moves are the legs of a right angled triangle. The distance d is its hypotenuse.
d² = 5² + 5² = 25 + 25 = 50.
d = √50 = 7.071 on the calculator. To 1 decimal place that is 7.1.

Answer: The huts are 7.1 km apart, to 1 decimal place.

A box has a longest stick too. It runs from a bottom corner to the opposite top corner. That line is the space diagonal. We find it with two right angled triangles, one flat on the floor and one standing up.

The box is drawn from the side, so the depth looks shorter than it is. The labels give the true lengths.

Triangle one finds x. Triangle two, built on x, finds d.

Worked example

A box is 4 cm long, 3 cm wide and 12 cm high. How long is the longest stick that fits inside?

The floor is a rectangle 4 cm by 3 cm. Its diagonal x is the hypotenuse of a right angled triangle.
x² = 4² + 3² = 16 + 9 = 25.
x = √25 = 5 cm.
Now stand a triangle on that diagonal. Its legs are x and the height, and the stick d is its hypotenuse.
d² = 5² + 12² = 25 + 144 = 169.
d = √169 = 13 cm.

Answer: The longest stick is 13 cm.

Look at the two sums together: d² = x² + 12², and x² = 4² + 3². So d² = 4² + 3² + 12². For any box, d² = l² + w² + h². You can go straight to it, but you must understand where it comes from.

See it three ways

With things

Lean a pencil in a box so it touches a bottom corner and the opposite top corner. It is the longest pencil the box holds.

As a picture

In symbols

d² = l² + w² + h², so d = √(l² + w² + h²).

Try it

Measure a diagonal, then check it

You need: a sheet of paper, a ruler.

Measure the long side and the short side of the sheet. Say they are 30 cm and 20 cm.
Work out 30² + 20², then take the square root.
Measure the diagonal, corner to corner. Compare.

What you will see: For 30 cm and 20 cm, the diagonal is √1 300, which is 36.06 cm. To the nearest centimetre it is 36 cm. Your measure should be close to the sum.

If you cannot try it, this is what happens.

A wrong idea many people have

The distance between two points is the across move plus the up move.

Why it feels right: On a grid you count squares along the lines, and counting is adding.

The test that breaks it: For A (1, 2) and B (7, 10), across plus up is 6 + 8 = 14. Measure A to B with the squares of the grid: it is 10 squares, so 10 km.

What is true: Across plus up is the way along the two legs. The straight line is the hypotenuse, which is shorter: d = √(6² + 8²) = 10.

A wrong idea many people have

The longest stick in a box is as long as its longest edge.

Why it feels right: A stick lying along the longest edge looks as long as the box allows.

The test that breaks it: In the 4 cm by 3 cm by 12 cm box, the longest edge is 12 cm. But a 13 cm stick fits when you slant it corner to corner, because √169 = 13.

What is true: The corner to corner line through the middle is longer than any edge. d = √(l² + w² + h²).

Key words

space diagonal: the line from one corner of a box to the opposite corner, through the inside.

Back to your guess

The answer is C: 7 cm, from one bottom corner to the opposite top corner.

The longest stick runs corner to corner through the middle, and it is 7 cm. The sum is 2² + 3² + 6² = 4 + 9 + 36 = 49, and √49 = 7. A stick cannot bend, so 11 cm is too long. This chapter uses right angled triangles to find such lengths.

Keep it

Copy this into your exercise book.

Spot the right angled triangle. A ladder is the hypotenuse. Across and up on a grid are the legs. In a box, d² = l² + w² + h².

Draw: A box with its floor diagonal x dashed and its space diagonal d heavy, and d² = l² + w² + h² under it.

M4.3.3 Chapter 3

Opposite, adjacent and hypotenuse

Two right angled triangles have the same angles, but one is twice as big. Is opposite ÷ adjacent bigger in the big one?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A No, it is the same, because both sides are multiplied by the same number.
You picked A. Read on. At the end, the chapter comes back to your guess.
B Yes, the big triangle has bigger sides, so the answer is bigger.
You picked B. Read on. At the end, the chapter comes back to your guess.
C No, it is smaller, because dividing makes things smaller.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Name the opposite, adjacent and hypotenuse for a given angle
Show by drawing that opposite ÷ adjacent is fixed for each angle

In Chapter 1 the sides had names that did not depend on an angle. Now we pick one of the two acute angles and call it θ, the Greek letter theta. The names of two sides then depend on where θ is.

The hypotenuse is still the side opposite the right angle. The opposite side faces θ: it is across from θ and does not touch it. The adjacent side is next to θ: it touches θ and is not the hypotenuse.

Draw it first

Name the sides for the angle θ

1 Pick an acute angle and mark it θ. The right angle has its small square.
2 Name the sides: hypotenuse faces the right angle, opposite faces θ, adjacent touches θ.

Choose the other angle and the names swap. The side that was opposite becomes adjacent, and the other way round. The hypotenuse never changes.

The same triangle. θ is at B now, so opposite and adjacent swap.

Worked example

In triangle PQR the right angle is at Q. PQ is 5 cm, QR is 12 cm and PR is 13 cm. The angle θ is at P. Name the sides and find opposite ÷ adjacent.

The right angle is at Q. The hypotenuse faces it: PR = 13 cm.
θ is at P. The opposite side is across from P: QR = 12 cm.
The adjacent side touches P and is not the hypotenuse: PQ = 5 cm.
Opposite ÷ adjacent = 12 ÷ 5 = 2.4.

Answer: Opposite ÷ adjacent = 12 ÷ 5 = 2.4.

Now the big idea. Draw an angle θ. Stand a vertical line at 4 cm, at 8 cm and at 12 cm along the flat arm. You get three right angled triangles. They all have the same θ.

One angle θ, three triangles. Each is the one before, made bigger.

The last column never changes.

Why? Triangle 2 is triangle 1 enlarged by a factor of 2: every side is doubled. Triangle 3 is an enlargement by a factor of 3. When all sides are multiplied by the same number, a division of two sides stays the same. So for one angle, opposite ÷ adjacent is a fixed number. Change the angle and the number changes.

We can use the fixed number. If we know θ gives 0.75, then in any right angled triangle with that θ, the opposite side is 0.75 times the adjacent side.

Worked example

A small triangle has opposite 3 cm and adjacent 4 cm for θ. A bigger triangle has the same θ and adjacent 12 cm. Find its opposite side x.

In the small triangle, opposite ÷ adjacent = 3 ÷ 4 = 0.75.
The big triangle has the same angle θ, so its opposite ÷ adjacent is also 0.75.
So x ÷ 12 = 0.75.
Multiply both sides by 12: x = 0.75 × 12 = 9.

Answer: The opposite side is 9 cm.

See it three ways

With things

Two ramps with the same steepness. The short one goes 4 m along and rises 3 m. The long one goes 8 m along and rises 6 m.

As a picture

In symbols

3 ÷ 4 = 6 ÷ 8 = 0.75.

Try it

Draw two triangles with one angle

You need: paper, a ruler, a protractor, a pencil.

Draw a flat line 10 cm long. At its left end, use the protractor to draw a line at 40° above it.
Stand a vertical line at 5 cm and another at 10 cm along the flat line, up to the slanted line.
Measure the two vertical lines. Divide each by its flat length.

What you will see: The vertical lines measure about 4.2 cm and 8.4 cm. Both divisions give about 0.84. A bigger triangle, the same answer.

If you cannot try it, this is what happens.

A wrong idea many people have

A bigger triangle with the same angle has a bigger opposite ÷ adjacent.

Why it feels right: Both the opposite side and the adjacent side get longer, and we think a longer side means a bigger answer.

The test that breaks it: Take 3 ÷ 4 = 0.75 and 6 ÷ 8 = 0.75. The two sides doubled, and the answer stayed at 0.75.

What is true: Doubling both sides leaves the division the same. Opposite ÷ adjacent depends only on the angle.

A wrong idea many people have

The adjacent side is always the one along the bottom.

Why it feels right: In most pictures the flat side is called adjacent, and a position is quicker to remember than a rule.

The test that breaks it: Turn the triangle so another side is on the bottom. Opposite and adjacent have not moved: they still face and touch θ.

What is true: Opposite and adjacent depend on where θ is, not on how the page is turned. Look for the side that touches θ and is not the hypotenuse.

Key words

opposite: the side across from the angle θ, not touching it.
adjacent: the side next to the angle θ that is not the hypotenuse.

Back to your guess

The answer is A: No, it is the same, because both sides are multiplied by the same number.

The answer is the same. In the big triangle both sides are twice as long, and 6 ÷ 8 = 3 ÷ 4. For a given angle, opposite ÷ adjacent never changes. That fixed number is the start of trigonometry.

Keep it

Copy this into your exercise book.

Hypotenuse faces the right angle. Opposite faces θ. Adjacent touches θ. For one angle, opposite ÷ adjacent is the same in every size of triangle.

Draw: A right angled triangle with θ marked, the three sides named, and a bigger copy inside it with the same θ.

M4.3.4 Chapter 4

Sine, cosine and tangent

A ladder 5 m long leans on a wall at 75° to the ground. How high up the wall does it reach?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A About 5 m, because the ladder is 5 m long.
You picked A. Read on. At the end, the chapter comes back to your guess.
B About 3.75 m, because 75% of 5 m is 3.75 m.
You picked B. Read on. At the end, the chapter comes back to your guess.
C About 4.8 m, because the ladder is nearly upright.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Choose sin, cos or tan from the sides you know
Use sin, cos or tan to find a missing side

In the last chapter you saw that opposite ÷ adjacent is fixed for an angle θ. That fixed number has a name. It is the tangent of θ, written tan θ.

There are two more fixed numbers, and the same argument shows they are fixed too. The sine of θ is opposite ÷ hypotenuse, written sin θ. The cosine of θ is adjacent ÷ hypotenuse, written cos θ.

The sides are 3, 4 and 5 for this θ.

Three ratios of the same triangle. Each is a fixed number for this θ.

For this triangle θ is about 36.9°. Put your calculator in degrees mode, often shown as DEG. Then sin 36.87° gives about 0.6, cos 36.87° gives about 0.8 and tan 36.87° gives about 0.75. The calculator holds a table of these numbers for every angle.

To remember the three ratios, use SOH CAH TOA. SOH: Sine is Opposite over Hypotenuse. CAH: Cosine is Adjacent over Hypotenuse. TOA: Tangent is Opposite over Adjacent.

To find a missing side, do three things. Label the sides O, A and H for your angle. Find the two sides that matter: the one you know and the one you want. Pick the ratio that has exactly those two letters.

Start with the ladder from the question. It makes a right angled triangle with the wall and the ground. The ladder is H, 5 m. The height h is opposite the 75° angle. The two letters are O and H, so you use sin.

Worked example

A ladder 5 m long leans on a wall at 75° to the ground. How high up the wall does it reach? Give the answer to 1 decimal place.

The ladder is the hypotenuse. The height h is opposite the 75° angle. We have O and H, so use sin.
sin 75° = h ÷ 5, so h = 5 × sin 75°.
On the calculator, to 4 decimal places, sin 75° = 0.9659.
Multiply: 5 × 0.9659 = 4.8295.
Round to 1 decimal place: h = 4.8 m.

Answer: To 1 decimal place, the ladder reaches 4.8 m.

Now a case with cosine. A ramp is 6 m long and slopes at 20° to the flat ground. How far does it stretch along the ground? Draw the triangle first.

Draw it first

Choose the ratio for a ramp

1 Draw the ramp as the slanted side. Mark the 20° angle with the ground.
2 Label the ramp 6 m, the hypotenuse. Label the flat distance x. It is adjacent to the 20° angle. We have A and H, so use cos.

Worked example

A ramp 6 m long slopes at 20° to flat ground. How far does it stretch along the ground? Give the answer to 1 decimal place.

The ramp is the hypotenuse. The distance x is adjacent to the 20° angle. We have A and H, so use cos.
cos 20° = x ÷ 6, so x = 6 × cos 20°.
On the calculator, to 4 decimal places, cos 20° = 0.9397.
Multiply: 6 × 0.9397 = 5.6382.
Round to 1 decimal place: x = 5.6 m.

Answer: To 1 decimal place, the ramp stretches 5.6 m.

Sometimes the side you want is on the bottom of the ratio. Then you divide. A tree is 12 m tall. From a point on level ground, the top is seen at an angle of 35° above the ground. How far is the point from the foot of the tree?

Worked example

The top of a tree 12 m tall is seen at 35° above the ground. How far is the viewer from the foot of the tree? Give the answer to 1 decimal place.

The 12 m height is opposite the 35° angle. The distance x is adjacent. We have O and A, so use tan.
tan 35° = 12 ÷ x, so x × tan 35° = 12.
Divide both sides by tan 35°: x = 12 ÷ tan 35°.
On the calculator, to 4 decimal places, tan 35° = 0.7002.
Divide, to 3 decimal places: 12 ÷ 0.7002 = 17.138.
Round to 1 decimal place: x = 17.1 m.

Answer: To 1 decimal place, the viewer is 17.1 m from the tree.

Keep the rule: when the unknown is on top of the ratio, multiply. When it is on the bottom, divide. In real work keep all the digits on your calculator and round only at the end. In the three examples we rounded the ratio to 4 places only to show each step. The final answers are the same.

See it three ways

With things

A ramp is 5 m long and rises 3 m. For every metre of ramp, you rise 3 ÷ 5 = 0.6 m.

As a picture

In symbols

sin θ = opposite ÷ hypotenuse = 3 ÷ 5 = 0.6.

Try it

Measure a sine

You need: paper, a ruler, a protractor, a pencil.

Draw a flat line. At its left end use the protractor to draw a line at 30° above it, 10 cm long. This is the hypotenuse.
From the top end of the slanted line, draw a line straight down to the flat line. Measure it.
Divide the measured length by 10.

What you will see: The vertical line is about 5 cm, so the sine is 5 ÷ 10 = 0.5. Your calculator says sin 30° = 0.5. The paper and the calculator agree.

If you cannot try it, this is what happens.

A wrong idea many people have

On my calculator sin 30° is −0.988, so the calculator has a fault.

Why it feels right: A calculator is exact, so a strange answer must be a broken button.

The test that breaks it: Look for DEG or D on the screen. Some calculators start in radians. In radians mode, sin 30 means 30 radians, and that gives −0.988.

What is true: Set the calculator to degrees. Then sin 30° = 0.5, which matches the drawing.

A wrong idea many people have

If sin 30° = 0.5, then sin 60° = 1, because 60° is twice 30°.

Why it feels right: Doubling the angle looks like it should double the answer.

The test that breaks it: Draw a 60° angle with a hypotenuse of 10 cm. The upright side is about 8.7 cm, so sin 60° is about 0.87, not 1.

What is true: The ratios do not grow in step with the angle. Use the calculator for each angle.

Key words

sine: sin θ = opposite ÷ hypotenuse.
cosine: cos θ = adjacent ÷ hypotenuse.
tangent: tan θ = opposite ÷ adjacent.

Back to your guess

The answer is C: About 4.8 m, because the ladder is nearly upright.

The ladder reaches about 4.8 m. It cannot reach 5 m, because the ladder is the hypotenuse and the height is a shorter side. A percentage of the angle is not a length. You find the height with 5 × sin 75°, and this chapter shows how.

Keep it

Copy this into your exercise book.

SOH CAH TOA: sin = O ÷ H, cos = A ÷ H, tan = O ÷ A. Label O, A, H. Pick the ratio with your two sides. Unknown on top: multiply. Unknown below: divide.

Draw: A right angled triangle with θ, the sides marked O, A and H, and the words SOH CAH TOA under it.

M4.3.5 Chapter 5

Finding angles

A ramp rises 1 m while it goes 4 m along the ground. What angle does it make with the ground?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A About 14°, found with the inverse tangent of 1 ÷ 4.
You picked A. Read on. At the end, the chapter comes back to your guess.
B 25°, because 1 ÷ 4 = 0.25, and the ratio is the angle.
You picked B. Read on. At the end, the chapter comes back to your guess.
C 45°, because the ramp goes up as far as it goes along.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Use inverse sin, cos or tan to find an angle
Find the angle of a slope from its height and length

In the last chapter the angle was given and a side was missing. Now it is the other way round. You know two sides and you want the angle.

You already know sin 30° = 0.5. Turn it round. Which angle has a sine of 0.5? The answer is 30°. The key that does this is the inverse sine, written sin⁻¹. On most calculators you press shift, then sin.

Reading the table from the right gives the inverse: which angle has sin 0.5? The row 30°.

Look at the table. As the angle grows, sin grows from 0 up to 1. A sine of 0.5 sits at 30°, and a sine of 0.87 at 60°. The inverse keys do the same search for any number, not only the ones in a small table.

Take care with the name. sin⁻¹ does not mean 1 ÷ sin. The small −1 says "go backwards". You find an angle from a ratio, and the answer is an angle in degrees. Check that the calculator is in degrees mode.

The three inverse keys follow the same rule as before. Choose the ratio from the two sides you know. Divide them. Then press the inverse key for that ratio.

A sine or cosine in a right angled triangle is always between 0 and 1, because the hypotenuse is the longest side. If your ratio is above 1, the sides are the wrong way round. The calculator shows an error for sin⁻¹ or cos⁻¹ of a number above 1. A tangent can be any number above 0.

Worked example

A ramp rises 0.5 m over a flat distance of 6 m. What angle does it make with the ground? Give the answer to 1 decimal place.

The rise 0.5 m is opposite θ. The flat 6 m is adjacent. We have O and A, so use tan.
tan θ = 0.5 ÷ 6.
Use the inverse tan key on 0.5 ÷ 6, without rounding the division first.
To 4 decimal places, θ = 4.7636.
Round to 1 decimal place: θ = 4.8°.

Answer: To 1 decimal place, the angle is 4.8°.

Now a road. A road is 150 m long and climbs 20 m. The road is the hypotenuse, so the angle uses sine.

Worked example

A road is 150 m long and climbs 20 m. What angle does the road make with the flat? Give the answer to 1 decimal place.

The climb 20 m is opposite θ. The road, 150 m, is the hypotenuse. We have O and H, so use sin.
To 4 decimal places, sin θ = 20 ÷ 150 = 0.1333.
Use the inverse sin key on 20 ÷ 150.
To 4 decimal places, θ = 7.6623.
Round to 1 decimal place: θ = 7.7°.

Answer: To 1 decimal place, the angle is 7.7°.

Last, a ladder. Its foot is 1.8 m from the wall and it is 6 m long. What angle does it make with the ground? Build the picture in three moves.

Draw it first

Find the angle of a ladder

1 Draw the ladder, the ground and the wall as a right angled triangle.
2 Write the known lengths: the foot distance 1.8 m and the ladder 6 m.
3 Mark θ where the ladder meets the ground.

Worked example

A ladder 6 m long has its foot 1.8 m from a wall. What angle does it make with the ground? Give the answer to 1 decimal place.

The foot distance 1.8 m is adjacent to θ. The ladder, 6 m, is the hypotenuse. We have A and H, so use cos.
cos θ = 1.8 ÷ 6 = 0.3.
Use the inverse cos key on 0.3.
To 4 decimal places, θ = 72.5424.
Round to 1 decimal place: θ = 72.5°.

Answer: To 1 decimal place, the angle is 72.5°.

See it three ways

With things

A road climbs 20 m for every 150 m you walk along it.

As a picture

In symbols

sin θ = 20 ÷ 150, so θ = sin⁻¹(20 ÷ 150), which is 7.7° to 1 decimal place.

Try it

Find the angle of a ruler

You need: a ruler 30 cm long, a pile of books, a protractor.

Make a pile of books 12 cm high. Lean the ruler from the floor to the top of the pile.
The ruler is 30 cm long and the pile is 12 cm high. Work out sin⁻¹(12 ÷ 30).
Now measure the angle between the ruler and the floor with the protractor.

What you will see: The inverse sine gives 23.6°, and the protractor should show about the same.

If you cannot try it, this is what happens.

A wrong idea many people have

The ratio is the angle: tan θ = 0.25 means θ = 25°.

Why it feels right: The digits 25 match, and a calculator shows a number, so the number seems to be the answer.

The test that breaks it: Work out tan 25° on the calculator. It gives 0.466, not 0.25. The angle that has tan 0.25 is 14.04°.

What is true: A ratio is a number about the shape, and an angle is a turn. Use the inverse key to change one into the other.

A wrong idea many people have

sin⁻¹ 0.5 means 1 ÷ sin 0.5.

Why it feels right: In algebra a small −1 above a number means the reciprocal: 2⁻¹ = 1 ÷ 2.

The test that breaks it: We know sin 30° = 0.5. If sin⁻¹ 0.5 were 1 ÷ sin 0.5, the answer would not be 30. The inverse key gives 30°, so it goes backwards, it does not divide.

What is true: On a function key, ⁻¹ means the inverse function. sin⁻¹ undoes sin.

Key words

inverse sine: sin⁻¹: the angle whose sine is a given ratio. Likewise cos⁻¹ and tan⁻¹.

Back to your guess

The answer is A: About 14°, found with the inverse tangent of 1 ÷ 4.

The angle is about 14°. The ratio 0.25 is not the angle. It is the tangent of the angle, and the inverse tan key on a calculator turns it back into 14.04°. A ramp at 45° would rise 4 m over 4 m, not 1 m.

Keep it

Copy this into your exercise book.

Two sides known, angle wanted. Pick sin, cos or tan. Divide the sides, then press the inverse key. Use degrees mode.

Draw: A right angled triangle with two sides labelled, θ marked, and θ = tan⁻¹(opposite ÷ adjacent) under it.

M4.3.6 Chapter 6

Exact values

On a calculator sin 45° shows 0.7071067812 and goes on. Is there a neat exact way to write it?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A No. The decimal never ends, so it cannot be written exactly.
You picked A. Read on. At the end, the chapter comes back to your guess.
B Yes. It is exactly 0.7071, the calculator is only adding extra digits.
You picked B. Read on. At the end, the chapter comes back to your guess.
C Yes. It is 1 ÷ √2, and half a square shows why.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Recall sin, cos and tan of 30°, 45° and 60°
Work out those values from two special triangles

A calculator gives sin 60° as 0.8660254... The digits never end and never repeat, so no decimal is exact. But for 30°, 45° and 60° we can write the exact value, as a fraction or with a square root. We get them from two triangles that take a minute to draw.

First triangle. Draw a square with sides of 1 cm. Cut it along a diagonal. You get two equal right angled triangles.

The diagonal is √2 cm because 1² + 1² = 2.

The diagonal cuts each right angle in half. So each triangle has angles 45°, 45° and 90°. Its sides are 1 cm, 1 cm and √2 cm.

Worked example

Use half a square with sides 1 cm to find sin 45°, cos 45° and tan 45° exactly.

Draw a square with sides 1 cm and cut it along a diagonal.
Each half has a right angle, two angles of 45° and two sides of 1 cm.
The hypotenuse by Pythagoras: c² = 1² + 1² = 2, so c = √2 cm.
For a 45° angle the opposite side is 1 and the hypotenuse is √2, so sin 45° = 1 ÷ √2.
The adjacent side is 1, so cos 45° = 1 ÷ √2 and tan 45° = 1 ÷ 1 = 1.

Answer: sin 45° = 1 ÷ √2, cos 45° = 1 ÷ √2 and tan 45° = 1.

You may meet √2 ÷ 2 instead of 1 ÷ √2. They are the same number. Multiply the top and the bottom of 1 ÷ √2 by √2, and you get √2 ÷ 2. Both are 0.7071... Many books prefer √2 ÷ 2 because it has no root on the bottom.

Second triangle. Draw an equilateral triangle with sides 2 cm. Cut it down the middle from the top corner to the middle of the base.

The cut line is √3 cm because 2² − 1² = 3.

The cut makes two right angled triangles. The top angle of 60° is split into two 30° angles. Each triangle has angles 30°, 60° and 90°, and sides 1 cm, √3 cm and 2 cm.

Worked example

Use half an equilateral triangle with sides 2 cm to find sin 30°, cos 30° and tan 30° exactly.

Draw an equilateral triangle with sides 2 cm and cut it in half down the middle.
Each half has a right angle and angles of 30° and 60°. Its hypotenuse is 2 cm and its short side is 1 cm.
The third side by Pythagoras: h² = 2² − 1² = 3, so h = √3 cm.
For the 30° angle the opposite side is 1 and the hypotenuse is 2, so sin 30° = 1/2.
The adjacent side is √3, so cos 30° = √3 ÷ 2 and tan 30° = 1 ÷ √3.

Answer: sin 30° = 1/2, cos 30° = √3 ÷ 2 and tan 30° = 1 ÷ √3.

The same triangle gives the 60° values. The 60° angle is at the other end, so opposite and adjacent swap. Then sin 60° = √3 ÷ 2, cos 60° = 1/2 and tan 60° = √3 ÷ 1 = √3.

Learn this table, or learn how to draw the two triangles and read it off.

In the table a slash is a division. So 1/√3 means 1 ÷ √3, and √3/2 means √3 ÷ 2. The two ways of writing say the same thing.

Exact values keep an answer exact. A wire 10 m long holds a pole upright. It makes an angle of 60° with the ground. How high up the pole is it fixed? Leave the answer with a root first, then round it.

Worked example

A wire 10 m long holds a pole upright and makes an angle of 60° with the ground. How high up the pole is it fixed? Give it exactly, then to 1 decimal place.

The height is opposite the 60° angle. The wire is the hypotenuse. We have O and H, so use sin.
sin 60° = √3 ÷ 2, so h = 10 × √3 ÷ 2.
Work out 10 ÷ 2 = 5.
So h = 5 × √3 = 5√3 m, exactly.
On the calculator, 5 × 1.7321 = 8.6605.
To 1 decimal place, h = 8.7 m.

Answer: The wire is fixed exactly 5√3 m up the pole, which is 8.7 m to 1 decimal place.

See it three ways

With things

Fold a square sheet of paper corner to corner. The fold is longer than a side, and the two halves are the same.

As a picture

In symbols

sin 45° = 1 ÷ √2, which is about 0.7071.

Try it

Fold a square

You need: a square sheet of paper, a ruler.

Cut a square with sides of 10 cm. Fold it corner to corner.
Measure the fold, which is the hypotenuse of each half.
Divide a side, 10 cm, by the fold.

What you will see: The fold is about 14.1 cm. The division 10 ÷ 14.1 is about 0.71. This is sin 45° = 1 ÷ √2, about 0.7071, as in the table.

If you cannot try it, this is what happens.

A wrong idea many people have

cos 30° is 1/2, because sin 30° is 1/2 and the two go together.

Why it feels right: The numbers 1/2 and √3 ÷ 2 appear together in the table, and the two are often swapped.

The test that breaks it: Draw the 30° triangle. The side next to the 30° angle is the long leg, √3 cm, not the short one. So cos 30° = √3 ÷ 2, about 0.87. The calculator agrees.

What is true: sin 30° = 1/2 and cos 30° = √3 ÷ 2. Cos 60° is the one that equals 1/2. The sine and the cosine of 30° and 60° swap.

A wrong idea many people have

1 ÷ √2 and √2 ÷ 2 are two different numbers.

Why it feels right: They look different, and one has a root on the bottom.

The test that breaks it: Multiply the top and the bottom of 1 ÷ √2 by √2. The top becomes 1 × √2 = √2. The bottom becomes √2 × √2 = 2. So 1 ÷ √2 = √2 ÷ 2.

What is true: They are the same number, 0.7071... Writing a root off the bottom is a matter of habit, not of value.

Key words

exact value: a value written with a fraction or a root, not rounded to decimals.

Back to your guess

The answer is C: Yes. It is 1 ÷ √2, and half a square shows why.

It is exactly 1 ÷ √2. A decimal that never ends can still be written exactly with a root. Half a square gives sin 45°, and half an equilateral triangle gives the values for 30° and 60°. In this chapter you will see where each comes from.

Keep it

Copy this into your exercise book.

Half a square gives 45°: sides 1, 1, √2. Half an equilateral triangle gives 30° and 60°: sides 1, √3, 2. Read sin, cos and tan from these.

Draw: The two triangles with their sides and angles, and a small table of sin, cos and tan for 30°, 45° and 60°.

M4.3.7 Chapter 7

The sine rule and the cosine rule

A triangle has no right angle. You know two of its angles and one side. How can you find the other sides?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A Cut it into right angled triangles, and the work gives two rules that work for any triangle.
You picked A. Read on. At the end, the chapter comes back to your guess.
B You cannot. Sine, cosine and tangent only work with a right angle.
You picked B. Read on. At the end, the chapter comes back to your guess.
C Guess the right angle, then use Pythagoras.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Use the sine rule to find a side or an angle
Use the cosine rule when you know two sides and the angle between

Name a triangle ABC. Use a capital letter for each angle, and the small letter for the side opposite it. So side a faces angle A, side b faces angle B and side c faces angle C.

Each side is named after the corner it faces.

Draw the height h from C down to the side c. It makes two right angled triangles. In the left one, sin A = h ÷ b, so h = b × sin A. In the right one, sin B = h ÷ a, so h = a × sin B.

The height h is a side of two right angled triangles.

Both are h, so b × sin A = a × sin B. Divide both sides by sin A × sin B. On the left you are left with b ÷ sin B. On the right you are left with a ÷ sin A. So a ÷ sin A = b ÷ sin B. Cutting from another corner gives the same with c.

This is the sine rule: a ÷ sin A = b ÷ sin B = c ÷ sin C. Our height fell inside the triangle, because angles A and B were both acute. The rule is true for every triangle. A little later we see why it holds for an obtuse angle too.

Use it when you have a matching pair: an angle and the side opposite it. Then one more piece lets you find another pair. Here is a side.

Worked example

In triangle ABC, A = 40°, B = 65° and a = 8 cm. Find the side b, to 1 decimal place.

We know a and A, a matching pair, and B. We want b, opposite B. Use the sine rule.
a ÷ sin A = b ÷ sin B, so b = a × sin B ÷ sin A.
b = 8 × sin 65° ÷ sin 40°.
To 4 decimal places, sin 65° = 0.9063 and sin 40° = 0.6428.
To 4 decimal places, 8 × 0.9063 ÷ 0.6428 = 11.2794.
Round to 1 decimal place: b = 11.3 cm.

Answer: To 1 decimal place, b is 11.3 cm.

You can also find an angle. Turn the rule upside down: sin A ÷ a = sin B ÷ b. Then sin B = b × sin A ÷ a, and the inverse sin key gives B.

Worked example

In triangle ABC, A = 50°, a = 12 cm and b = 9 cm. Find the angle B, to 1 decimal place.

We know a and A, a matching pair, and b. We want B, opposite b. Use the sine rule upside down.
sin B ÷ b = sin A ÷ a, so sin B = b × sin A ÷ a.
sin B = 9 × sin 50° ÷ 12.
To 4 decimal places, sin B = 9 × 0.7660 ÷ 12 = 0.5745.
Use the inverse sin key on 9 × sin 50° ÷ 12, with all the digits. To 3 decimal places, B = 35.067.
Round to 1 decimal place: B = 35.1°.
Here b is shorter than a, so B is smaller than A and must be acute. 35.1° is the only answer.

Answer: To 1 decimal place, B is 35.1°.

So far every angle was below 90°. A triangle can have an angle above 90°, called an obtuse angle. Then the height from C falls outside the triangle, on the line c extended beyond A.

The height falls outside. The angle outside at A is 180° − 120° = 60°.

The right angled triangle outside has the angle 180° − A at A. In the picture that is 60°. Its height is h = b × sin 60°.

We want h = b × sin A to stay true for the obtuse angle A. So we choose sin A to mean sin(180° − A). It is a definition. We choose it so that the rules in this chapter keep working.

We make the same choice for the cosine: cos A = −cos(180° − A). Here is why. When A is obtuse, the foot of the height lies beyond A. So the far part of the base is c + b cos(180° − A). With our choice that is c − b cos A again, so the cosine rule keeps its form. Check on a calculator. sin 114.6° and sin 65.4° both give 0.9092. cos 100° gives −0.1736 and cos 80° gives 0.1736. A negative cosine for an angle above 90° is not a fault.

Take care. When you find an angle with the sine rule, there can be two answers. Two different angles have the same sine: an acute angle and 180° minus it. The calculator gives only the acute one.

Look at the picture. Both triangles have A = 30°, a = 5.5 cm and b = 10 cm. Swing the 5.5 cm side round C. It touches the base line in two places, B₁ and B₂. Both make real triangles: 30° + 65.4° and 30° + 114.6° are both under 180°.

The same three facts, two triangles. The tick marks CB₁ and CB₂ as equal. B is 65.4° at B₁ or 114.6° at B₂.

The two answers for B are 65.4° and 180° − 65.4° = 114.6°. Always test the second angle. Add it to the known angle: 30° + 114.6° = 144.6°, which is below 180°. So the second triangle exists.

There is at most one triangle in these cases. First, the angle you find is opposite the shorter of the two known sides. Then it must be acute. That was the case before: B faced b = 9 cm, shorter than a = 12 cm.

Second, the known angle is 90° or more. Then the side opposite it must be the longest side, or there is no triangle. Third, the known angle and the second angle add to 180° or more. If sin B works out above 1, there is no triangle at all.

The sine rule needs a matching pair. Sometimes you have no pair. You may know two sides and the angle between them, or all three sides. Then use the second rule.

Draw the height h from C onto side c again. It cuts c into two parts. The part next to A is b × cos A. The rest is c − b × cos A. The height is h = b × sin A.

The foot of the height cuts c into b cos A and c − b cos A.

Now use Pythagoras in the right angled triangle on the right: a² = h² + (c − b cos A)². Put in h = b sin A.

First expand the bracket. Multiply each part of (c − b cos A) by each part of the other bracket. Then (c − b cos A)(c − b cos A) = c² − 2bc cos A + b² cos²A.

So a² = b² sin²A + c² − 2bc cos A + b² cos²A. Collect the two b² terms. They give b² × (sin²A + cos²A).

In the left triangle Pythagoras says (b sin A)² + (b cos A)² = b². So sin²A + cos²A = 1, and the sum is b². Therefore a² = b² + c² − 2bc cos A.

This is the cosine rule: a² = b² + c² − 2bc cos A. Our height fell inside, but the rule holds for every triangle. For an obtuse A the cosine is negative, so −2bc cos A adds to the sum. If A = 90°, then cos A = 0 and the rule is just Pythagoras. The extra part −2bc cos A corrects for the angle not being 90°.

Worked example

In triangle ABC, b = 8 cm, c = 11 cm and A = 50°. Find the side a, to 1 decimal place.

We know two sides, b = 8 and c = 11, and the angle A = 50° between them. We want a. Use the cosine rule.
a² = b² + c² − 2 × b × c × cos A = 8² + 11² − 2 × 8 × 11 × cos 50°.
Work out the squares and the product: a² = 64 + 121 − 176 × cos 50°.
To 4 decimal places, cos 50° = 0.6428.
Then a² = 185 − 176 × 0.6428 = 71.8672.
Take the square root, to 3 decimal places: a = 8.477.
Round to 1 decimal place: a = 8.5 cm.

Answer: To 1 decimal place, a is 8.5 cm.

The cosine rule can also find an angle when you know all three sides. Rearrange it: cos A = (b² + c² − a²) ÷ (2 × b × c). Then use the inverse cos key.

Worked example

A triangle has sides a = 7 cm, b = 8 cm and c = 9 cm. Find the angle A, opposite a, to 1 decimal place.

We know all three sides. We want A, opposite a. Use the cosine rule.
Rearrange: cos A = (b² + c² − a²) ÷ (2 × b × c).
cos A = (8² + 9² − 7²) ÷ (2 × 8 × 9).
The top is 64 + 81 − 49 = 96.
The bottom is 2 × 8 × 9 = 144.
To 4 decimal places, cos A = 96 ÷ 144 = 0.6667.
Use the inverse cos key on 96 ÷ 144. To 4 decimal places, A = 48.1897.
Round to 1 decimal place: A = 48.2°.

Answer: To 1 decimal place, A is 48.2°.

See it three ways

With things

Two sticks joined at one end, 6 cm and 8 cm long, opened to a 60° angle. How far apart are the free ends?

As a picture

In symbols

a² = 6² + 8² − 2 × 6 × 8 × cos 60° = 36 + 64 − 48 = 52. So a = √52, which is 7.2 cm to 1 decimal place.

Try it

Two sticks and a string

You need: two sticks or pencils of 6 cm and 8 cm, a protractor, a ruler.

Lay the two sticks from one point, with a 60° angle between them. Use the protractor.
Measure the distance between the free ends.
Work out 6² + 8² − 2 × 6 × 8 × cos 60°, then take the square root.

What you will see: The sum gives √52, which is 7.2 cm to 1 decimal place. Your measure should be close.

If you cannot try it, this is what happens.

A wrong idea many people have

Use the sine rule or the cosine rule, whichever comes to mind first.

Why it feels right: Both rules have a side and an angle in them, so either seems to fit.

The test that breaks it: Try two sides and the angle between them. The sine rule has no matching pair, so you cannot start. With a pair, the sine rule needs just one more piece.

What is true: Pick the rule from what you know. A matching pair (an angle and its opposite side) means the sine rule. Two sides and the angle between, or three sides, means the cosine rule.

A wrong idea many people have

The sine rule always gives exactly one angle.

Why it feels right: The calculator shows one answer, and one answer looks final.

The test that breaks it: Take A = 30°, a = 5.5 cm and b = 10 cm. sin B = 0.909, so the calculator gives 65.4°. But 114.6° has the same sine, and a triangle with 30° and 114.6° is fine.

What is true: Sine gives two angles, B and 180° − B. The second fits only when the known angle plus 180° − B is under 180°. That needs B opposite the longer of the two known sides. If sin B works out above 1, no triangle has those measures.

Key words

sine rule: a ÷ sin A = b ÷ sin B = c ÷ sin C, true for every triangle.
cosine rule: a² = b² + c² − 2bc cos A, true for every triangle.

Back to your guess

The answer is A: Cut it into right angled triangles, and the work gives two rules that work for any triangle.

Cutting a triangle with a height makes two right angled triangles. That gives the sine rule and the cosine rule, which work in every triangle. This chapter builds both rules from the height, then uses them.

Keep it

Copy this into your exercise book.

Sine rule: a ÷ sin A = b ÷ sin B. Use it with a matching pair. Cosine rule: a² = b² + c² − 2bc cos A. Use it with two sides and the angle between, or three sides.

Draw: A triangle ABC with its height h dashed, and the two rules written under it.

M4.3.8 Chapter 8

Area of any triangle, and bearings

A walker leaves camp A and walks to B on a bearing of 065°. What bearing takes her back to A?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A 115°, because 180° − 65° = 115°.
You picked A. Read on. At the end, the chapter comes back to your guess.
B 245°, because turning round adds a half turn: 65° + 180° = 245°.
You picked B. Read on. At the end, the chapter comes back to your guess.
C 295°, because 360° − 65° = 295°.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Find the area of a triangle as 1/2 ab sin C
Give a direction as a three figure bearing, like 065°
Solve a journey in two parts with bearings and the cosine rule

You know that the area of a triangle is 1/2 × base × height. The height is not always given. Sometimes you know two sides and the angle between them. Then trigonometry finds the height for you.

Take a as the base. The height h stands on it.

In the right angled triangle at the right, sin C = h ÷ b, so h = b × sin C. Then area = 1/2 × a × h = 1/2 × a × b × sin C. This is the area rule: area = 1/2 ab sin C, where C is the angle between the sides a and b.

The angle must be the one between the two sides you use. Our height fell inside the triangle. If it falls outside, the sine of an obtuse angle gives the same height. The last chapter showed why, so the rule still holds.

Worked example

A triangle has sides a = 10 cm and b = 8 cm, and the angle between them is C = 50°. Find its area, to 1 decimal place.

The angle C = 50° is between the sides a = 10 and b = 8. Use area = 1/2 × a × b × sin C.
To 4 decimal places, sin 50° = 0.7660.
Area = 0.5 × 10 × 8 × 0.7660 = 30.64.
Round to 1 decimal place: area = 30.6 cm².

Answer: To 1 decimal place, the area is 30.6 cm².

Now bearings. A bearing gives a direction as an angle measured clockwise from north. It is always written with three figures, so 65° is written 065°. North is 000°, east is 090°, south is 180° and west is 270°.

Bearings turn clockwise, the way a clock hand turns.

To draw a bearing, stand at the start. Draw a line north. Turn clockwise by the angle. Here is 065° from A to B, 5 km.

Going back from B to A is the opposite direction, a half turn of 180°. The north lines at A and B are parallel, so the bearing at B is 065° + 180° = 245°. If the sum is 360° or more, take 360° away. For a bearing of 250°, 250° + 180° = 430°. Take 360° away to get 070°. That is the same as taking 180° away, which is quicker for any bearing of 180° or more.

Worked example

The bearing of B from A is 065°. Find the bearing of A from B.

A to B is a bearing of 065°. Going back from B to A turns the direction through a half turn, 180°.
The north lines at A and B are parallel, so the half turn works at B too.
Add 180°: 65 + 180 = 245.
The result is below 360, so write it with three figures: 245°.

Answer: The bearing of A from B is 245°.

Now a journey in two parts. A boat leaves A and sails 8 km on a bearing of 070° to B. It turns and sails 6 km on a bearing of 150° to C. How far is C from A, and on what bearing?

First find the angle ABC inside the triangle. At B the way back to A is the bearing 070° + 180° = 250°. The way on to C is 150°. The angle between the two lines is 250° − 150° = 100°.

The angle inside the triangle at B is the difference of the two bearings: 250° − 150° = 100°.

Worked example

A boat sails 8 km on a bearing of 070° from A to B. Then it sails 6 km on a bearing of 150° to C. How far is C from A, to 1 decimal place?

Draw the journey. A is the start, B is the turn and C is the end.
At B the bearing of A is 70 + 180 = 250°. The bearing of C is 150°.
The angle ABC is the difference: 250 − 150 = 100°.
We know two sides, 8 and 6, and the angle between them. Use the cosine rule: AC² = 8² + 6² − 2 × 8 × 6 × cos 100°.
To 4 decimal places, cos 100° = −0.1736. The cosine of an obtuse angle is negative, as the last chapter showed.
Use all the digits of cos 100°. To 4 decimal places, AC² = 116.6702.
To 4 decimal places, AC = 10.8014.
Round to 1 decimal place: AC = 10.8 km.

Answer: To 1 decimal place, C is 10.8 km from A.

Now the bearing of C from A. Find the angle BAC with the sine rule. Then add it to 070°, because from A the point C lies clockwise of B.

The bearing of C from A is the bearing of B, 070°, plus the angle BAC, 33.2°.

Worked example

Use the same journey. Find the bearing of C from A, to the nearest degree.

We want the angle BAC. BC = 6 is opposite it, and AC = 10.8014 is opposite the angle 100°. Use the sine rule.
sin A ÷ 6 = sin 100° ÷ 10.8014, so sin A = 6 × sin 100° ÷ 10.8014.
Use all the digits on your calculator. To 4 decimal places, sin A = 0.5470.
The inverse sin key gives, to 2 decimal places, A = 33.16.
BC is the shortest side, so A is acute and 33.16° is the only answer.
From A, the point B is on 070° and C is clockwise of B. The bearing of C is 70 + 33.16 = 103.16.
Round to the nearest degree: the bearing of C from A is 103°.

Answer: The bearing of C from A is 103°.

See it three ways

With things

A path leaves a hut and runs north east. North east is half way between north and east.

As a picture

In symbols

North east = half of 090° = 045°.

Try it

Draw a bearing

You need: paper, a ruler, a protractor.

Mark a point A. Draw a straight line up the page from A, 3 cm long. This is north.
Place the protractor at A. Measure 65° clockwise from the north line, and mark it.
Draw a line 5 cm long from A through the mark. Label the end B.

What you will see: The line AB points in the direction 065°. Now stand at B, draw a north line, and measure clockwise to A. You should find 245°.

If you cannot try it, this is what happens.

A wrong idea many people have

Bearings turn anticlockwise, like angles in a maths class.

Why it feels right: Angles in a graph are measured anticlockwise from the right, so it seems natural to do the same with directions.

The test that breaks it: West is 270° on a bearing. If bearings turned anticlockwise, a quarter turn from north would be west, 090°. The compass picture shows east is 090°.

What is true: A bearing starts at north and turns clockwise, the way a clock hand does, and it has three figures.

A wrong idea many people have

The area rule works with any angle in the triangle.

Why it feels right: There are three angles and sin C looks like a general part of the rule.

The test that breaks it: For sides 10 cm and 8 cm with 50° between them the area is 30.6 cm². Using the 51.6° angle at B, which is not between those sides, gives 31.4 cm². That is a different answer, and it is wrong.

What is true: The angle must be between the two sides in the formula. If you have sides a and b, use angle C. If you have b and c, use A.

Key words

area rule: area = 1/2 ab sin C, with C the angle between sides a and b.
bearing: a direction given as an angle clockwise from north, written with three figures.

Back to your guess

The answer is B: 245°, because turning round adds a half turn: 65° + 180° = 245°.

The bearing back is 245°. A bearing is the angle clockwise from north. Going back is a half turn, 180°, from the way you came, so you add 180°. This chapter explains bearings, and finds areas of triangles with no right angle.

Keep it

Copy this into your exercise book.

Area = half × a × b × sin C, with C between a and b. A bearing is clockwise from north, in three figures. Back bearing: add or take away 180°.

Draw: A triangle with the angle C between sides a and b. Beside it, a north line with a bearing arc.

Squares on the sides

A square corner from string

Find the triangle in the ladder

Measure a diagonal, then check it

Name the sides for the angle θ

Draw two triangles with one angle

Choose the ratio for a ramp

Measure a sine

Find the angle of a ladder

Find the angle of a ruler

Fold a square

Two sticks and a string

Draw a bearing

Key words, A to Z