Mini-Lesson

Trigonometry

3 missions. Guess, learn, then check. Your place is saved.

Mission 1Will a 1.2 m umbrella fit flat in a 1 m by 0.8 m box?
Mission 2How can you find a tree's height with a protractor and a tape?
Mission 3Two bearings, 10 km apart: how far away is the lighthouse?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

Will a 1.2 m umbrella fit flat in a 1 m by 0.8 m box?

Make a guess: Will a 1.2 m umbrella fit flat in a 1 m by 0.8 m box?

Geometry · Pythagoras' theorem

The hypotenuse and the theorem

In a right-angled triangle, the side opposite the right angle is the hypotenuse. It is always the longest side.
Pythagoras' theorem: if the shorter sides are $a$ a and $b$ b and the hypotenuse is $c$ c, then

a^{2} + b^{2} = c^{2}

The theorem is about the squares of the sides, not the sides themselves.
It works the other way too: sides 6, 8 and 10 give $36 + 64 = 100$ 36 + 64 = 100, so that triangle is right-angled. Sides 6, 8 and 11 do not.

Worked example Geometry · Pythagoras' theorem

Finding the hypotenuse

Question

A right-angled triangle has shorter sides 8 cm and 5 cm. Find the length of the hypotenuse to 3 significant figures.

1
Square the two shorter sides and add: $8^{2} + 5^{2} = 64 + 25 = 89$ 8² + 5² = 64 + 25 = 89
2
That is $c^{2}$ c², so square-root: $c = 89 = 9.433 \dots$ c = √89 = 9.433…
3
Round to 3 significant figures and check it is longer than 8 cm.

Answer

9.43 cm

⚠️ Watch out

stopping at 89 forgets the square root. A "hypotenuse" of 89 cm next to sides of 8 cm and 5 cm is impossible.

Calculate

?A right-angled triangle has shorter sides 9.5 cm and 6.2 cm. Find the length of the hypotenuse. Give your answer to 1 decimal place.

cm

Geometry · Pythagoras' theorem

Finding a shorter side

Rearrange the theorem: $a^{2} = c^{2} - b^{2}$ a² = c² − b².
Square, subtract, square-root. Subtract because the hypotenuse is the biggest square.
Check: your answer must be shorter than the hypotenuse.
First spot the hypotenuse: it is opposite the right angle, wherever the triangle is turned.

Worked example Geometry · Pythagoras' theorem

A ladder against a wall

Question

A 5 m ladder leans against a vertical wall. Its foot is 1.4 m from the wall on level ground. How high up the wall does it reach?

1
The ladder is the hypotenuse (opposite the right angle between wall and ground): $c = 5$ c = 5
2
Square and subtract: $h^{2} = 5^{2} - 1. 4^{2} = 25 - 1.96 = 23.04$ h² = 5² − 1.4² = 25 − 1.96 = 23.04
3
Square-root: $h = 23.04 = 4.8$ h = √23.04 = 4.8

Answer

4.8 m

⚠️ Watch out

adding the squares gives

26.96 \approx 5.19

m, which is longer than the ladder itself.

Quick check

?A right-angled triangle has hypotenuse 13.5 cm and one shorter side 8.1 cm. What is the length of the third side,

x

?

Calculate

?A rectangular field is 120 m long and 50 m wide. Sam walks from one corner to the opposite corner along the edges. Ali walks straight across the diagonal. How much shorter is Ali's walk?

m

Geometry · Pythagoras' theorem

Pythagoras in 3-D (Extended)

To find the longest diagonal of a cuboid, use Pythagoras twice.
First the diagonal across the base: $d^{2} = a^{2} + b^{2}$ d² = a² + b². Then the base diagonal and the height make a right angle: $g^{2} = d^{2} + c^{2}$ g² = d² + c².
Together: $g^{2} = a^{2} + b^{2} + c^{2}$ g² = a² + b² + c², so $g = a^{2} + b^{2} + c^{2}$ g = √(a² + b² + c²).
A box 1 m by 0.8 m by 0.5 m deep: $g = 1 + 0.64 + 0.25 = 1.89 \approx 1.37$ g = √(1 + 0.64 + 0.25) = √1.89 ≈ 1.37 m.

Calculate

?A box is a cuboid 60 cm long, 40 cm wide and 30 cm high. Find the length of the longest straight rod that fits inside it. Give your answer to 1 decimal place.

cm

Mission check

?Will a 1.2 m umbrella fit flat in a box 1 m by 0.8 m?

Mission 2

How can you find a tree's height with a protractor and a tape?

Make a guess: How can you find a tree's height with a protractor and a tape?

Geometry · trigonometry

Naming the sides and the three ratios

Mark the angle $θ$ θ you know or want. The hypotenuse is opposite the right angle, the opposite faces $θ$ θ, the adjacent is next to $θ$ θ (and not the hypotenuse).
$sin θ = \frac{opp}{hyp} cos θ = \frac{adj}{hyp} tan θ = \frac{opp}{adj}$ sin θ = opp/hyp cos θ = adj/hyp tan θ = opp/adj
Memory aid: SOH CAH TOA. Pick the ratio that uses the side you have and the side you want.
Each ratio is fixed for a given angle because all right-angled triangles with that angle are similar.

Worked example Geometry · trigonometry

Finding a side

Question

In a right-angled triangle the hypotenuse is 12 cm and one angle is 35°. Find the side opposite the 35° angle, to 3 significant figures.

1
Label from 35°: hyp = 12, opp = $x$ x. These are O and H, so use sine.
2
Write the ratio: $sin 3 5^{\circ} = \frac{x}{12}$ sin 35° = x/12
3
Multiply both sides by 12: $x = 12 sin 3 5^{\circ} = 6.883 \dots$ x = 12sin 35° = 6.883…

Answer

6.88 cm

⚠️ Watch out

check your calculator is in degrees. In radians

12 sin 35

gives

- 5.14

, which is impossible for a length.

Worked example Geometry · trigonometry

Finding an angle

Question

A right-angled triangle has opposite side 5 cm and adjacent side 8 cm from angle

θ

. Find

θ

to 1 decimal place.

1
O and A, so use tangent: $tan θ = \frac{5}{8} = 0.625$ tan θ = 5/8 = 0.625
2
Use the inverse function to undo tan: $θ = tan^{- 1} (0.625)$ θ = tan⁻¹(0.625)
3
Calculate: $θ = 32.005 \dots^{\circ}$ θ = 32.005…°

Answer

θ = 32. 0^{\circ}

⚠️ Watch out

tan θ = \frac{8}{5}

gives 58.0°, the other acute angle. Opposite goes on top.

Quick check

?In a right-angled triangle one angle is 40°. The side opposite it is 6 cm. The hypotenuse is

x

. Which calculation gives

x

?

Geometry · trigonometry

Angles of elevation and depression

The angle of elevation is measured up from the horizontal to the line of sight.
The angle of depression is measured down from the horizontal.
Looking up from A to B and down from B to A gives equal angles (alternate angles between parallel horizontals).
Measure from where your eye is: add the eye height back on at the end.
Extended: elevation and depression; also, the shortest distance from a point to a line is the perpendicular one.

Calculate

?Priya stands on level ground 25 m from the base of a tree. Her eyes are 1.6 m above the ground. The angle of elevation of the top of the tree from her eyes is 32°. Find the height of the tree, to 1 decimal place.

m

Calculate

?A ship sails 12 km from a harbour on a bearing of 065° (measured clockwise from north). How far east of the harbour is it? Give your answer to 3 significant figures.

km

Geometry · trigonometry

Exact values

Extended content.
Learn these (from half an equilateral triangle of side 2 and an isosceles right-angled triangle with shorter sides 1):

$θ$ θ	0°	30°	45°	60°	90°
$sin θ$ sin θ	0	$\frac{1}{2}$ 1/2	$\frac{2}{2}$ (√2)/2	$\frac{3}{2}$ (√3)/2	1
$cos θ$ cos θ	1	$\frac{3}{2}$ (√3)/2	$\frac{2}{2}$ (√2)/2	$\frac{1}{2}$ 1/2	0
$tan θ$ tan θ	0	$\frac{3}{3}$ (√3)/3	1	$3$ √3	—

$\frac{3}{3} = \frac{1}{3}$ (√3)/3 = 1/(√3) and $\frac{2}{2} = \frac{1}{2}$ (√2)/2 = 1/(√2). $tan 9 0^{\circ}$ tan 90° is undefined.
Non-calculator use: hypotenuse 10 and angle 30° give opposite $10 \times \frac{1}{2} = 5$ 10 × 1/2 = 5.

Worked example Geometry · trigonometry

The angle between a line and a plane (Extended)

Question

A cuboid ABCDEFGH has base ABCD with AB = 8 cm, BC = 6 cm, and height CG = 5 cm. Find the angle between the diagonal AG and the base ABCD, to 1 decimal place.

1
Drop from G to the base: it lands on C, so the angle is $∠ G A C$ ∠ GAC in triangle GAC (right angle at C).
2
Find AC across the base with Pythagoras: $A C = 8^{2} + 6^{2} = 10$ AC = √(8² + 6²) = 10
3
In triangle GAC, opp = CG = 5 and adj = AC = 10: $tan θ = \frac{5}{10}$ tan θ = 5/10
4
Inverse: $θ = tan^{- 1} (0.5) = 26.56 \dots^{\circ}$ θ = tan⁻¹(0.5) = 26.56…°

Answer

26.6°

⚠️ Watch out

using the edge AB = 8 instead of AC gives 32.0°. The angle with a plane uses the line's projection onto it.

Mission check

?Sol stands on level ground 20 m from a tree, eyes 1.5 m above the ground, and measures a 40° angle up to the top. How tall is the tree?

Mission 3

Two bearings, 10 km apart: how far away is the lighthouse?

Make a guess: Two bearings, 10 km apart: how far away is the lighthouse?

Geometry · non-right-angled triangles

The sine rule (Extended)

Label angles $A$ A, $B$ B, $C$ C and the opposite sides $a$ a, $b$ b, $c$ c.
Sine rule: each side divided by the sine of its opposite angle gives the same value.

\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} or \frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}

Use it when you know a side and its opposite angle; use the second form for an angle.
These formulas are given on the formula sheet.
Obtuse angles: $sin x = sin (18 0^{\circ} - x)$ sin x = sin (180° − x); $sin^{- 1}$ sin⁻¹ gives only the acute one.
Ambiguous case: $a = 7$ a = 7, $b = 9$ b = 9, $A = 4 0^{\circ}$ A = 40°, $sin B = \frac{9 s i n 4 0 ^{\circ}}{7}$ sin B = (9sin 40°)/7 gives $B = 55. 7^{\circ}$ B = 55.7° or $124. 3^{\circ}$ 124.3°; both fit: two triangles.

Worked example Geometry · non-right-angled triangles

A bearings problem with the sine rule (Extended)

Question

Boat B is 12 km due east of boat A. A buoy C is on a bearing of 050° from A and 300° from B. Find the distance BC, to 3 significant figures.

1
Angle at A: from east (090°) back to 050° is $9 0^{\circ} - 5 0^{\circ} = 4 0^{\circ}$ 90° − 50° = 40°
2
Angle at B: from west (270°) round to 300° is $3 0^{\circ}$ 30°, so $C = 18 0^{\circ} - 4 0^{\circ} - 3 0^{\circ} = 11 0^{\circ}$ C = 180° − 40° − 30° = 110°
3
Pair AB = 12 with its opposite angle $C = 11 0^{\circ}$ C = 110°; BC is opposite $A = 4 0^{\circ}$ A = 40°: $\frac{BC}{sin 4 0 ^{\circ}} = \frac{12}{sin 11 0 ^{\circ}}$ BC/(sin 40°) = 12/(sin 110°)
4
Rearrange: $BC = \frac{12 sin 4 0 ^{\circ}}{sin 11 0 ^{\circ}} = 8.208 \dots$ BC = (12sin 40°)/(sin 110°) = 8.208…

Answer

BC = 8.21 km

⚠️ Watch out

using

sin 3 0^{\circ}

on top pairs BC with the wrong angle and gives 6.39 km, which is AC.

Calculate

?In triangle ABC,

a = 9

cm,

b = 7

cm and angle

A = 7 2^{\circ}

. Find angle

B

, to 1 decimal place.

degrees

Geometry · non-right-angled triangles

The cosine rule (Extended)

Use the cosine rule when you know two sides and the angle between them, or all three sides.

a^{2} = b^{2} + c^{2} - 2 b c cos A cos A = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c}

$a$ a is the side opposite the angle $A$ A; $b$ b and $c$ c are the sides that make angle $A$ A.
If $A = 9 0^{\circ}$ A = 90°, $cos A = 0$ cos A = 0 and it becomes Pythagoras.
For an obtuse angle $cos A$ cos A is negative, so the calculator gives the obtuse angle directly.

Worked example Geometry · non-right-angled triangles

Finding an angle from three sides (Extended)

Question

A triangle has sides 5 cm, 7 cm and 10 cm. Find its largest angle, to 1 decimal place.

1
The largest angle is opposite the longest side, so $a = 10$ a = 10, $b = 5$ b = 5, $c = 7$ c = 7
2
Substitute: $cos A = \frac{5 ^{2} + 7 ^{2} - 1 0 ^{2}}{2 \times 5 \times 7} = \frac{- 26}{70}$ cos A = (5² + 7² − 10²)/(2 × 5 × 7) = (−26)/70
3
Evaluate: $cos A = - 0.3714 \dots$ cos A = −0.3714…
4
Inverse cosine: $A = 111.80 \dots^{\circ}$ A = 111.80…°

Answer

111. 8^{\circ}

⚠️ Watch out

the negative cosine is correct, not a mistake: it tells you the angle is obtuse.

Calculate

?In triangle ABC,

b = 8

cm,

c = 11

cm and the angle between them is

A = 3 8^{\circ}

. Find side

a

, to 3 significant figures.

cm

Sort it

?Which rule do you start with?

Tap a statement, then tap the group it belongs in. Then press Check.

Geometry · non-right-angled triangles

Area of any triangle (Extended)

Two sides and the angle between them give the area:

Area = \frac{1}{2} ab sin C

The angle must be the included angle, between sides $a$ a and $b$ b.
Example: sides 6 cm and 9 cm with 30° between them: $\frac{1}{2} \times 6 \times 9 \times sin 3 0^{\circ} = 13.5 cm^{2}$ 1/2 × 6 × 9 × sin 30° = 13.5 cm².
If you know all three sides, find an angle with the cosine rule first.

Calculate

?A triangular garden has sides 8 m, 11 m and 13 m. Find its area, to 3 significant figures.

m²

Mission check

?A ship at A sees a lighthouse L on a bearing of 040°. It sails 10 km due north to B, where L is on a bearing of 120°. How far is the ship at B from the lighthouse?

Recap

Big ideas

The hypotenuse is opposite the right angle and is the longest side: $a^{2} + b^{2} = c^{2}$ a² + b² = c².
Hypotenuse: square, add, square-root. Shorter side: square, subtract, square-root.
Extended: in a cuboid use Pythagoras twice, $g = a^{2} + b^{2} + c^{2}$ g = √(a² + b² + c²).
Label hyp, opp and adj from the angle; choose the ratio with SOH CAH TOA.
Unknown side: write the ratio and rearrange. Unknown angle: use $sin^{- 1}$ sin⁻¹, $cos^{- 1}$ cos⁻¹ or $tan^{- 1}$ tan⁻¹.
Know the exact values for 0°, 30°, 45°, 60°, 90°; in 3-D, find a right-angled triangle inside the solid (Higher/Extended).
Label each side with the same letter as its opposite angle.
Sine rule for a side–angle pair; cosine rule for two sides with the included angle, or three sides.
Area $= \frac{1}{2} ab sin C$ = 1/2 absin C with the angle between the two sides.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Trigonometry for Cambridge IGCSE Maths (0580). 🎉

Mission 1Will a 1.2 m umbrella fit flat in a 1 m by 0.8 m box?
Mission 2How can you find a tree's height with a protractor and a tape?
Mission 3Two bearings, 10 km apart: how far away is the lighthouse?

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Specification points and sources for this lesson are listed at the bottom of the page.