Relatively
Puzzled.

M3b Trigonometry from scratch

use sine, cosine and tangent confidently on right-angled triangles in any orientation.

Before you begin: M3a · Angles and radians.

Videos coming soon

One 22-minute video and a 5-minute worked example (how tall is that tree?) are being made, and will appear here once they're uploaded to YouTube.

77 steps about 78 min

Watch first

Notes available, video soon.

Notes

In this chapter

  • Use similar triangles: same angles, same shape, same ratios.
  • Use Pythagoras' theorem, \(a^2 + b^2 = c^2\), and see why it's true.
  • Name the hypotenuse, opposite and adjacent sides from any angle, in any orientation.
  • Use sine, cosine and tangent to find sides, and the inverse functions to find angles.
  • Know the special angles 30°, 45° and 60°, and \(\sin^2\theta + \cos^2\theta = 1\).
  • Turn gradients into angles, and measure heights you can't reach (with a home-made clinometer).

Similar triangles

FIRST, MAKE A PREDICTION

Two right triangles have the same acute angle. One is three times as large. What happens to opposite side divided by hypotenuse?

Show the answer

The ratio stays the same

Both lengths scale by three, so the scale factor cancels in their ratio.

TRY IT / FOLLOW THE REASONING

Keep the model's angle fixed and change the hypotenuse. Compare the side lengths and the opposite-to-hypotenuse ratio, then change the angle instead.

Explore a right triangle

Angle: 30 ° · Hypotenuse: 5 m

Opposite
2.5 m
Adjacent
4.33 m
Opposite ÷ hypotenuse
0.5
Model assumptions

A right triangle; the angle is measured from the horizontal. Lengths are in metres. The drawing fits the available space; use the readouts to compare lengths.

MAKE THE CONNECTION

Key idea

Triangles with the same three angles are similar: one is a scaled copy of the other, so the ratios of their sides are equal.

The Sun's rays are parallel, so a tree and a stick make similar shadow triangles:

Similar triangles

\[ \frac{\text{tree height}}{\text{tree shadow}} = \frac{\text{stick height}}{\text{stick shadow}} \]

Similar triangles

Two similar shadow triangles sharing the Sun angle of 32.0 degrees: a 1 m stick with a 1.6 m shadow inside a tree 7.5 m tall with a 12 m shadow.stick 1 m, shadow 1.6 m32.0°tree shadow 12 mtree 7.5 mSun's ray
The tree and the stick make similar triangles with the Sun's ray: the same angle, 32.0° up, and the same shape, so the same ratios.

Similar triangles

Worked number: the tree

A 1 m stick with a 1.6 m shadow, beside a tree with a 12 m shadow: the tree is \(12 \div 1.6 = 7.5\) m tall. (Thales is said to have measured a pyramid this way, in the 6th century BC.)

Checkpoint: Similar triangles (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A 0.50 m stick casts a 0.80 m shadow. At the same moment a building casts a 40 m shadow. How tall is the building, in metres?

m

Show the answer

25 m

The ratio height ÷ shadow is the same for both: \(0.50/0.80 = 0.625\), so the building is \(40 \times 0.625 = 25\) m tall.

Checkpoint: Similar triangles (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

Two triangles have the same three angles, but one is three times as big as the other. What is true of their sides?

Show the answer

Their sides are in the same ratios, so the ratio of any two sides is the same in both.

The same three angles mean the same shape. One triangle is a scaled copy of the other, so the ratios of its sides match.

Pythagoras' theorem

FIRST, MAKE A PREDICTION

The perpendicular sides of a right triangle are 3 m and 4 m. What is its hypotenuse?

Show the answer

5 m

The square of the hypotenuse is 3² + 4² = 25, so its length is 5 m.

TRY IT / FOLLOW THE REASONING

Draw squares on the three sides of a 3–4–5 triangle. Compare their areas and identify why the right angle is essential to this relationship.

MAKE THE CONNECTION

Key idea

In a right-angled triangle, the longest side, across from the right angle, is the hypotenuse \(c\). The squares on the other two sides add up to the square on the hypotenuse.

Pythagoras' theorem

\[ \begin{gathered} a^2 + b^2 = c^2 \\[.6em] \text{e.g. } 3^2 + 4^2 = 9 + 16 = 25 = 5^2 \end{gathered} \]

Why: four copies of the triangle fit inside a square of side \(a + b\) in two ways. One leaves a tilted square \(c^2\) uncovered; the other leaves two squares, \(a^2\) and \(b^2\). Same square, same triangles, same leftover area.

Pythagoras' theorem

Two squares of side 7, each holding four copies of a triangle with sides 3, 4 and 5. The left leaves one tilted square of area 25; the right leaves squares of area 9 and 16.c²a²b²four triangles and c²four triangles, a² and b²
The same big square, filled two ways. Take away the four identical triangles from each and what is left must match: c² = a² + b², here 25 = 9 + 16.

Pythagoras' theorem

Worked number: the ladder

A 5 m ladder with its foot 1.4 m from a wall reaches \(\sqrt{25 - 1.96} = \sqrt{23.04} = 4.8\) m up the wall.

Checkpoint: Pythagoras' theorem (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A rectangular screen is 40 cm wide and 30 cm tall. How long is its diagonal, in centimetres?

cm

Show the answer

50 cm

\(c = \sqrt{40^2 + 30^2} = \sqrt{2500} = 50\) cm, a 3, 4, 5 triangle ten times bigger.

Checkpoint: Pythagoras' theorem (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

In a right-angled triangle, which side is the hypotenuse?

Show the answer

The longest side, across from the right angle.

The hypotenuse is across from the right angle, and it is the longest side, so \(c^2 = a^2 + b^2\).

Naming the sides

FIRST, MAKE A PREDICTION

You switch from one acute angle of a right triangle to the other. What happens to the side names?

Show the answer

Opposite and adjacent swap; the hypotenuse stays the same

Opposite and adjacent depend on the chosen angle. The hypotenuse is always across from the right angle.

TRY IT / FOLLOW THE REASONING

Draw one right triangle and label the sides from its lower acute angle. Repeat from its upper acute angle without rotating the page.

MAKE THE CONNECTION

Key idea

The names follow the angle, not the page. Pick an angle \(\theta\) (not the right angle): the opposite side is across from \(\theta\); the adjacent side touches \(\theta\) and isn't the hypotenuse.

Naming the sides

A right-angled triangle with the angle θ, about 36.9 degrees, at the left. The side across from θ is the opposite, 3; the side touching θ is the adjacent, 4; the longest side, across from the right angle, is the hypotenuse, 5.θadjacent 4opposite 3hypotenuse 5
The three sides of a 3, 4, 5 triangle, named from the angle θ: the opposite is across from it, the adjacent touches it, and the hypotenuse is across from the right angle.

Naming the sides

Use the other angle and opposite and adjacent swap. Turning or flipping the triangle changes nothing. Routine: find the right angle, then the hypotenuse across from it, then the side across from \(\theta\).

Checkpoint: Naming the sides (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

You pick an angle \(\theta\) in a right-angled triangle. What is the side that touches \(\theta\) but is not the hypotenuse called?

Show the answer

The adjacent side.

Opposite is across from \(\theta\); adjacent touches \(\theta\) and is not the hypotenuse.

Checkpoint: Naming the sides (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

In a right-angled triangle you switch from one acute angle to the other. What happens to the opposite and adjacent sides?

Show the answer

They swap: each one becomes the other.

The names follow the angle you chose. Use the other angle and opposite and adjacent swap; the hypotenuse stays put.

Sine, cosine and tangent

FIRST, MAKE A PREDICTION

In a right triangle, the side opposite θ is 3 and the hypotenuse is 5. Which ratio equals sin θ?

Show the answer

3/5

Sine is opposite divided by hypotenuse, so the ratio is 0.6.

TRY IT / FOLLOW THE REASONING

Change the model's angle and watch the two legs change while the hypotenuse stays fixed. Check which leg approaches zero as the angle becomes small.

Explore a right triangle

Angle: 30 ° · Hypotenuse: 5 m

Opposite
2.5 m
Adjacent
4.33 m
Opposite ÷ hypotenuse
0.5
Model assumptions

A right triangle; the angle is measured from the horizontal. Lengths are in metres. The drawing fits the available space; use the readouts to compare lengths.

MAKE THE CONNECTION

Key idea

Because similar triangles have equal ratios, each angle has its own fixed ratios. Sine, cosine and tangent are just names for them.

Sine, cosine and tangent

\[ \begin{gathered} \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} \\[.6em] \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} \\[.6em] \tan\theta = \frac{\text{opposite}}{\text{adjacent}} \end{gathered} \]

Rearranged: opposite = hypotenuse × sin θ, adjacent = hypotenuse × cos θ, opposite = adjacent × tan θ.

Sine, cosine and tangent

Worked number: ladder and ramp

A 5 m ladder at 70° reaches \(5\sin 70^\circ \approx 4.70\) m up the wall, with its foot \(5\cos 70^\circ \approx 1.71\) m out.

A 4 m ramp at 10° rises \(4\sin 10^\circ \approx 0.69\) m.

Sine, cosine and tangent

A 5 m ladder leaning on a wall at 70 degrees to the ground: its foot is 1.71 m out and it reaches 4.7 m up the wall.70°1.71 m out4.70 m up5 m ladder
The 5 m ladder at 70°: the side across from the angle is 5 sin 70° = 4.70 m, the side along the ground is 5 cos 70° = 1.71 m.

Sine, cosine and tangent

\(\theta\) 0° 30° 45° 60° 90°
\(\sin\theta\) 0 1/2 \(1/\sqrt2 \approx 0.707\) \(\sqrt3/2 \approx 0.866\) 1
\(\cos\theta\) 1 \(\sqrt3/2 \approx 0.866\) \(1/\sqrt2 \approx 0.707\) 1/2 0
\(\tan\theta\) 0 \(1/\sqrt3 \approx 0.577\) 1 \(\sqrt3 \approx 1.732\) —

(From half a square, sides 1, 1, \(\sqrt2\); and half an equilateral triangle, sides 1, \(\sqrt3\), 2.)

On a unit circle (radius 1), the point at angle \(\theta\) is at \((\cos\theta, \sin\theta)\), so Pythagoras gives \(\sin^2\theta + \cos^2\theta = 1\) for every angle. For small angles in radians, \(\sin\theta \approx \theta\) (M3a's small-angle rule).

Sine, cosine and tangent

The sine and cosine of an angle from 0 to 360 degrees. Both swing between −1 and 1; sine starts at 0, cosine at 1. Marked: sin 30° = 0.5, which is also cos 60°.90180270360−1.5−1−0.50.511.5sin θcos θθ (degrees)
Sine and cosine from 0° to 360°: both stay between −1 and 1, and sin 30° = 0.5 (the dot), which is also cos 60°.

Checkpoint: Sine, cosine and tangent (1 of 3)

Question 1 of 3 Commit to an answer first, then look.

A ramp 8.0 m long makes an angle of 15° with the ground. How high does it rise, in metres?

m

Show the answer

2.07055236082 m

opposite = hypotenuse \(\times \sin\theta = 8.0 \sin 15^\circ \approx 2.07\) m.

Checkpoint: Sine, cosine and tangent (2 of 3)

Question 2 of 3 Commit to an answer first, then look.

A 10 m rope is pulled tight from the top of a pole to the ground, at 60° to the ground. How far is its foot from the pole, in metres?

m

Show the answer

5 m

adjacent = hypotenuse \(\times \cos\theta = 10 \cos 60^\circ = 5\) m, half the rope.

Checkpoint: Sine, cosine and tangent (3 of 3)

Question 3 of 3 Commit to an answer first, then look.

An angle has \(\sin\theta = 0.6\). Using \(\sin^2\theta + \cos^2\theta = 1\), what is \(\cos\theta\)? (Take the positive value.)

Show the answer

0.8

\(\cos^2\theta = 1 - 0.36 = 0.64\), so \(\cos\theta = 0.8\): the 3, 4, 5 triangle again.

Going backwards: inverse functions

FIRST, MAKE A PREDICTION

What does tan⁻¹(0.625) ask a calculator to find?

Show the answer

The angle whose tangent is 0.625

An inverse trigonometric function recovers an angle from a ratio.

TRY IT / FOLLOW THE REASONING

Find an angle from a known opposite-to-adjacent ratio, then put that angle back through tangent. Check your degree or radian mode before comparing.

MAKE THE CONNECTION

Key idea

\(\theta = \tan^{-1}(0.625)\) means "the angle whose tangent is 0.625". The −1 is a name, not "one over".

Going backwards: inverse functions

Worked number: the Sun's height

A 1 m stick with a 1.6 m shadow means the Sun is \(\tan^{-1}(1/1.6) \approx 32.0^\circ\) above the horizon.

Going backwards: inverse functions

Calculator walkthrough

  • Check the screen shows D or DEG. Then sin 30 = gives 0.5.
  • Backwards: SHIFT (or 2nd), sin, 0.5, = gives 30.
  • The stick: SHIFT, tan, (, 1 ÷ 1.6, ), = gives 32.005…
  • The trap: in RAD mode, sin 30 gives −0.988 (30 radians is almost five turns). Quick check: sin 90 must give exactly 1.

Checkpoint: Going backwards: inverse functions (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

An angle has a sine of 0.8. What is the angle, in degrees?

°

Show the answer

53.1301023542 °

\(\theta = \sin^{-1}(0.8) \approx 53.1^\circ\). Check: \(\sin 53.1^\circ \approx 0.8\).

Checkpoint: Going backwards: inverse functions (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

What does \(\tan^{-1}(0.5)\) mean?

Show the answer

The angle whose tangent is 0.5.

\(\tan^{-1}(0.5)\) is the angle whose tangent is 0.5, about 26.6°.

Gradients and heights

FIRST, MAKE A PREDICTION

A road has a 100% gradient. What is its angle above the horizontal?

Show the answer

45°

A 100% gradient means rise equals horizontal run, so tan θ = 1.

TRY IT / FOLLOW THE REASONING

Sketch roads with rises of 1 and 10 for a horizontal run of 10. Calculate their gradients, then use inverse tangent to compare the actual angles.

MAKE THE CONNECTION

Key idea

A road's gradient is rise ÷ run × 100% = \(\tan\theta \times 100\%\). A 10% hill is only \(\tan^{-1}(0.1) \approx 5.7^\circ\); 100% is 45°.

Real data: Baldwin Street, Dunedin, the world's steepest street at 34.8% (measured along its centre line), is about 19.2°.

Gradients and heights

Baldwin Street: a gradient of 34.8%, a rise of 34.8 m for every 100 m along, which is an angle of 19.2 degrees.19.2°run 100 mrise 34.8 m
A gradient of 34.8% is a rise of 34.8 m for every 100 m along: an angle of 19.2°.

Gradients and heights

The angle of elevation is how far you tilt up from level to see the top of something. Then height = distance × tan(elevation), plus your eye height. Measuring a baseline and angles to fix distant points is triangulation: from 200 m away, a tower whose top is 25° up is \(200\tan 25^\circ \approx 93.3\) m tall (above your eye).

Gradients and heights

Try it at home: build a clinometer

Tape a straw along a protractor's straight edge, tie string through the centre hole, and hang a small weight on it. Sight the top of a tree through the straw; the string hangs straight down. The angle of elevation is 90° minus the reading: a reading of 55° means 35°.

Fifteen metres from the tree, that's \(15\tan 35^\circ + 1.6 \approx 12.1\) m (with an eye height of 1.6 m).

Gradients and heights

Sighting a tree from 15 m away at an elevation of 35 degrees: it rises 10.5 m above your eye, so with an eye height of 1.6 m it is 12.1 m tall.35°15 m to the tree10.5 m above the eye
From 15 m away at 35° elevation, the tree rises 10.5 m above your eye.

Checkpoint: Gradients and heights (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A path has a gradient of 20% (a rise of 0.20 for every 1 along the ground). What angle does it make with the level, in degrees?

°

Show the answer

11.309932474 °

\(\theta = \tan^{-1}(0.20) \approx 11.3^\circ\): about half of 20.

Checkpoint: Gradients and heights (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

You stand 50 m from a tower and its top is 30° above the horizontal. How tall is the tower above your eye, in metres?

m

Show the answer

28.8675134595 m

height = distance \(\times \tan\theta = 50 \tan 30^\circ \approx 28.9\) m.

Trig mistakes

FIRST, MAKE A PREDICTION

You enter sin(30) expecting 0.5 but get about −0.988. What should you check first?

Show the answer

Whether the calculator is in radian mode

sin(30°) is 0.5, but sin(30 radians) is about −0.988.

TRY IT / FOLLOW THE REASONING

Test a known right angle in each calculator mode. Write the unit beside the angle before evaluating, then check a triangle's opposite and hypotenuse labels.

MAKE THE CONNECTION

Myths and mistakes

  • Calculator in radian mode: sin 30 gives −0.988. Check that sin 90 gives 1.
  • Opposite and adjacent swapped: label from the angle you're using.
  • \(\tan^{-1}x\) means \(1/\tan x\): no, it's the angle whose tangent is \(x\).
  • A 10% hill is 10°: it's about 5.7°.
  • A bigger triangle has a bigger sine: no, the ratios depend only on the angle.

Checkpoint: Trig mistakes (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

You type sin 90 and the calculator shows 0.894. What is the most likely cause?

Show the answer

The calculator is in radian mode.

In radian mode sin 90 is 0.894. The quick check: in degree mode, sin 90 gives exactly 1.

Checkpoint: Trig mistakes (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

A road has a gradient of 10%. About what angle is that?

Show the answer

About 5.7°.

A gradient of 10% means \(\tan\theta = 0.1\), so \(\theta \approx 5.7^\circ\). A 100% gradient is 45°.

Key terms

Key terms

Similar triangles
Triangles with the same angles; their sides are in the same ratios.
Hypotenuse / opposite / adjacent
Across from the right angle / across from θ / next to θ (not the hypotenuse).
Sine, cosine, tangent
Opposite ÷ hypotenuse, adjacent ÷ hypotenuse, opposite ÷ adjacent.
Inverse functions
\(\sin^{-1}\), \(\cos^{-1}\), \(\tan^{-1}\): from a ratio back to an angle.

Key terms

Unit circle
A circle of radius 1; the point at angle θ is \((\cos\theta, \sin\theta)\).
Gradient
Rise ÷ run (× 100%) = \(\tan\theta\).
Angle of elevation, triangulation, clinometer
The angle up from level to a point; fixing distant points from a baseline and angles; an instrument for measuring angles of slope or elevation.

History and sources

Thales and the pyramid (6th century BC, as told later)

Thales and the pyramid (6th century BC, as told later). Hieronymus (quoted by Diogenes Laertius) says Thales measured a pyramid by its shadow "at the moment when our shadows are equal to our own height"; Plutarch's later version uses a stick and two similar triangles. Source: MacTutor: Thales.

Plimpton 322 (Babylon, about 1800 BC)

Plimpton 322 (Babylon, about 1800 BC). A clay tablet listing right-angled triangles with whole-number sides (first row 119, 120, 169), more than a thousand years before Pythagoras, whose own role is uncertain. The oldest surviving proof is Euclid's Elements I.47 (about 300 BC). Sources: Plimpton 322, Pythagorean theorem: history.

Hipparchus's table of chords (2nd century BC)

Hipparchus's table of chords (2nd century BC). The first known trigonometric table; his books are lost, and we know of them through Theon of Alexandria and Ptolemy. Source: MacTutor: Hipparchus.

The word "sine" (499 to the 12th century)

The word "sine" (499 to the 12th century). Aryabhata's half-chord ardha-jya (half a bow-string) became Arabic jiba, written jb; twelfth-century translators read it as jaib (fold) and wrote Latin sinus. Source: Sine and cosine: etymology.

The HP-35 (California, 1972)

The HP-35 (California, 1972). The first pocket calculator with trigonometric functions, built to fit Bill Hewlett's shirt pocket; $395. Source: HP-35.

The Great Trigonometrical Survey and Peak XV (1802–1856)

The Great Trigonometrical Survey and Peak XV (1802–1856). Peak XV was sighted from six stations over 100 km away; the computations were led by Radhanath Sikdar; the height announced in 1856 was 29,002 ft (8,840 m). The 2020 survey gives 8,848.86 m. Sources: Great Trigonometrical Survey, Radhanath Sikdar.

The world's steepest street (2019–2020)

The world's steepest street (2019–2020). Source: Baldwin Street.

Ladder angle:

Ladder angle: "The correct angle for a ladder is 75 degrees or the 1 in 4 rule" (Health and Safety Authority, Ireland: Using Ladders Safely).

Worked example: how tall is that tree?

The problem

Problem. (Illustrative measurements.) A 1.00 m stick casts a 1.19 m shadow; at the same moment, a tree's shadow reaches 14.6 m from the middle of its trunk. Standing 15.0 m from the tree, a clinometer's string reads 55°, with an eye height of 1.6 m. How tall is the tree, measured both ways?

Solution

  1. Shadows: similar triangles: \(h = 14.6 \times 1.00 \div 1.19 \approx 12.3\) m. (The Sun is \(\tan^{-1}(1/1.19) \approx 40.0^\circ\) up; check: \(14.6\tan 40.0^\circ \approx 12.3\) m.)
  2. Clinometer: elevation \(= 90^\circ - 55^\circ = 35^\circ\); \(15.0 \tan 35^\circ \approx 10.5\) m above the eye; add 1.6 m: about 12.1 m.
  3. Compare: they differ by about 1.4%: the tree is about 12 m tall.

Solution

  1. Errors: a 1° slip in the clinometer reading (34° or 36°) gives 11.7 m or 12.5 m: about ±0.4 m. The shadow method depends on finding the middle of the trunk, level ground, and the highest twig being above the trunk.

Solution

  1. Where to stand: the error is smallest at about 45°, so stand about as far away as the tree is tall.

Practice

★ Warm-up

  1. A right-angled triangle has shorter sides 6 and 8. Find the hypotenuse, and the sine, cosine and tangent of the angle opposite the side of 6. What is that angle, in degrees?

    °

    Solution

    \(c = \sqrt{36 + 64} = 10\). \(\sin\theta = 6/10 = 0.6\), \(\cos\theta = 8/10 = 0.8\), \(\tan\theta = 6/8 = 0.75\). \(\theta = \tan^{-1}(0.75) \approx 36.9^\circ\).

Problem 2 (★ Warm-up)

  1. Find \(\sin 35^\circ\), \(\cos 35^\circ\) and \(\tan 35^\circ\). Then find the angle whose sine is 0.5, and the angle whose tangent is 2.

    Solution

    0.574, 0.819, 0.700. \(\sin^{-1}(0.5) = 30^\circ\); \(\tan^{-1}(2) \approx 63.4^\circ\).

Problem 3 (★ Warm-up)

  1. Your calculator says \(\sin 90 = 0.894\). What has gone wrong?

    Solution

    It's in radian mode: 0.894 is the sine of 90 radians. Switch to degrees, and \(\sin 90 = 1\).

★★ Standard

  1. A ladder. A 6 m ladder stands with its foot 1.5 m out from a wall (a quarter of its length). What angle does it make with the ground, in degrees? Then work out how high up the wall it reaches.

    °

    Solution

    The ladder is the hypotenuse; 1.5 m is adjacent to the ground angle: \(\cos\theta = 1.5/6 = 0.25\), \(\theta \approx 75.5^\circ\): about the 75° that ladder-safety guides recommend. Height \(= \sqrt{36 - 2.25} \approx 5.81\) m.

Problem 5 (★★ Standard)

  1. A ramp. A ramp has a gradient of 1 in 12 (rising 1 for every 12 along the ground). If it rises 0.5 m, how long is the ramp itself, in metres?

    m

    Solution

    \(\tan^{-1}(1/12) \approx 4.76^\circ\). Run \(= 0.5 \times 12 = 6\) m. Length \(= \sqrt{6^2 + 0.5^2} \approx 6.02\) m.

Problem 6 (★★ Standard)

  1. A hill. You drive 400 m along a road with a 15% gradient. How much height do you gain, in metres?

    Hint

    The 400 m is along the road: the hypotenuse. Find the angle first.

    m

    Solution

    \(\theta = \tan^{-1}(0.15) \approx 8.53^\circ\). Height \(= 400\sin 8.53^\circ \approx 59.3\) m.

Problem 7 (★★ Standard)

  1. A flagpole. The shadow of a flagpole is 9.0 m long when the Sun is 38° above the horizon. How tall is the flagpole, in metres?

    m

    Solution

    The shadow is adjacent to the Sun's angle and the pole is opposite: \(h = 9.0\tan 38^\circ \approx 7.03\) m.

Problem 8 (★★ Standard)

  1. The Sun's height. A 1.00 m stick casts a 1.19 m shadow. How high is the Sun above the horizon, in degrees?

    °

    Solution

    \(\tan\theta = 1.00/1.19\), so \(\theta = \tan^{-1}(0.840) \approx 40.0^\circ\).

Problem 9 (★★ Standard)

  1. A steep street. A street has a gradient of 12%. What angle does it make with the level, in degrees?

    °

    Solution

    \(\tan\theta = 0.12\), so \(\theta = \tan^{-1}(0.12) \approx 6.84^\circ\): far less than 12°.

★★★ Stretch

  1. Height from two angles. From a point on level ground, the top of a tower is 30° above the horizontal. You walk 50 m straight towards it, and now it's 45° up. How tall is the tower (above your eye), in metres?

    Hint

    Call the second distance \(d\). Then \(h = d\tan 45^\circ\) and \(h = (d + 50)\tan 30^\circ\). Set them equal.

    m

    Solution

    \(d = (d + 50)\tan 30^\circ \Rightarrow d = \dfrac{50\tan 30^\circ}{1 - \tan 30^\circ} \approx 68.3\) m, so \(h \approx 68.3\) m, and you started \(68.3 + 50 = 118.3\) m away. This is how the surveyors could measure a mountain without knowing its distance first.

Problem 11 (★★★ Stretch)

  1. Extension: no right angle. The cosine rule, \(c^2 = a^2 + b^2 - 2ab\cos C\), works in any triangle. Two sides of 7 and 9 meet at 40°. How long is the third side? What does the rule become when \(C = 90^\circ\)?

    Solution

    \(c^2 = 49 + 81 - 126\cos 40^\circ \approx 33.5\), so \(c \approx 5.79\). At 90°, \(\cos C = 0\) and it becomes Pythagoras.

Problem 12 (★★★ Stretch)

  1. How small is small? Compare \(\sin 10^\circ\) with 10° in radians. By what percentage does the small-angle rule overestimate?

    Solution

    \(\sin 10^\circ \approx 0.17365\); \(10^\circ = 0.17453\) rad: about 0.51% too big.

Check yourself

Answer these out loud, from memory. Then try again in a few days.

Check yourself

  • Why do similar triangles have equal side ratios?

Check yourself

  • What does Pythagoras' theorem say, and why is it true?

Check yourself

  • How do you find the opposite and adjacent sides for a given angle, in any orientation?

Check yourself

  • What are sine, cosine and tangent, and why do they depend only on the angle?

Check yourself

  • What does \(\tan^{-1}\) mean, and how do you check your calculator's mode?

Check yourself

  • How is a road gradient related to its angle?

Check yourself

  • How can you find a height you can't reach?

Check yourself

  • How do angles, radians, triangles, and trigonometric functions enter physical problems?

Summary

Summary

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Next: Vectors and components

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Further learning

Khan Academy (found in Khan Academy's search index on 2026-10-07; Khan Academy blocks automated link checking, so tell us if one has moved):

Free textbook (checked 2026-10-07): OpenStax, Algebra and Trigonometry 2e, 7.2 Right Triangle Trigonometry (sine, cosine, tangent, special angles and angles of elevation).

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