Chapter 3: Right-Angled Trigonometry
Trigonometry extends Pythagoras by allowing you to work with angles as well as sides. In a right-angled triangle the three ratios — sine, cosine, and tangent — relate every angle to two of its sides. Mastering SOHCAHTOA unlocks a vast range of problems: from finding the height of a building using a single distance measurement, to navigating using bearings, to finding angles inside 3D solids.
Exam Board Coverage
Sections 3.1–3.5 appear at both Foundation and Extended tier. Section 3.4 (Exact Values) and Section 3.7 (3D Trigonometry) are primarily Extended tier topics. Always set your calculator to Degree mode before starting a trigonometry question.
3.1 Labelling Sides: O, A, H
Before choosing a trigonometric ratio, you must label the three sides of the right-angled triangle relative to the angle you are using.
Side Labels
- H — Hypotenuse: the side opposite the right angle (always the longest side).
- O — Opposite: the side directly opposite the angle $\theta$ that you are working with.
- A — Adjacent: the side next to the angle $\theta$ that is not the hypotenuse.
These labels change depending on which angle you choose. Always re-label when you switch to a different angle.
Right-angled triangle: angle $\theta$ at vertex $A$. Side $H$ (hypotenuse), side $O$ (opposite $\theta$), side $A$ (adjacent to $\theta$).
3.2 The Three Ratios: SOHCAHTOA
SOHCAHTOA
For an angle $\theta$ in a right-angled triangle:
$$\sin\theta = \frac{O}{H} \qquad \cos\theta = \frac{A}{H} \qquad \tan\theta = \frac{O}{A}$$
The mnemonic SOH–CAH–TOA stands for:
- Sine = Opposite over Hypotenuse
- Cosine = Adjacent over Hypotenuse
- Tangent = Opposite over Adjacent
Choosing the Right Ratio
Label all three sides O, A, H for the chosen angle. Then identify which two sides are involved in the question (the given side and the unknown side). The ratio that uses exactly those two sides is the one to use.
Ratio Selection Guide
- O and H involved → use sin
- A and H involved → use cos
- O and A involved → use tan
3.3 Finding a Side
If an angle and one side are known, you can find any other side using a trigonometric ratio.
Method: Finding an Unknown Side
- Label O, A, H for the given angle.
- Identify which two sides are involved; choose the appropriate ratio.
- Write the equation: e.g. $\sin\theta = \dfrac{O}{H}$.
- Substitute the known values.
- Rearrange to find the unknown side and evaluate.
Example 3.1 — Finding the opposite side
In a right-angled triangle, angle $\theta = 35°$ and the hypotenuse is $12\,\text{cm}$. Find the side opposite $\theta$.
Step 1 Label: H = 12, O = ?, A = (not needed). O and H are involved → use sin.
Step 2 $\sin 35° = \dfrac{O}{12}$
Step 3 $O = 12 \sin 35° = 12 \times 0.5736\ldots \approx \mathbf{6.88\,\text{cm}}$ (3 s.f.)
Example 3.2 — Finding the hypotenuse
The adjacent side to a $42°$ angle is $9\,\text{m}$. Find the hypotenuse.
Step 1 A = 9, H = ?, A and H involved → use cos.
Step 2 $\cos 42° = \dfrac{9}{H}$
Step 3 $H = \dfrac{9}{\cos 42°} = \dfrac{9}{0.7431\ldots} \approx \mathbf{12.1\,\text{m}}$ (3 s.f.)
Example 3.3 — Using tangent
Find the side adjacent to a $58°$ angle if the opposite side is $15\,\text{cm}$.
Step 1 O = 15, A = ?, O and A → use tan.
Step 2 $\tan 58° = \dfrac{15}{A}$
Step 3 $A = \dfrac{15}{\tan 58°} = \dfrac{15}{1.6003\ldots} \approx \mathbf{9.37\,\text{cm}}$ (3 s.f.)
3.4 Finding an Angle
When two sides are known, you can find an angle using the inverse trigonometric functions: $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$ (also written $\arcsin$, $\arccos$, $\arctan$).
Inverse Trigonometric Functions
$$\theta = \sin^{-1}\!\left(\frac{O}{H}\right) \qquad \theta = \cos^{-1}\!\left(\frac{A}{H}\right) \qquad \theta = \tan^{-1}\!\left(\frac{O}{A}\right)$$
On a calculator: press SHIFT (or 2nd) then sin, cos, or tan.
Example 3.4 — Finding an angle using sine
In a right-angled triangle the opposite side is $7\,\text{cm}$ and the hypotenuse is $11\,\text{cm}$. Find angle $\theta$.
Step 1 O = 7, H = 11 → use sin.
Step 2 $\sin\theta = \dfrac{7}{11}$
Step 3 $\theta = \sin^{-1}\!\left(\dfrac{7}{11}\right) \approx \mathbf{39.5°}$ (1 d.p.)
Example 3.5 — Finding an angle using cosine
A right-angled triangle has adjacent side $8\,\text{cm}$ and hypotenuse $14\,\text{cm}$. Find the angle.
Step 1 A = 8, H = 14 → use cos.
Step 2 $\cos\theta = \dfrac{8}{14} = \dfrac{4}{7}$
Step 3 $\theta = \cos^{-1}\!\left(\dfrac{4}{7}\right) \approx \mathbf{55.4°}$ (1 d.p.)
3.5 Exact Trigonometric Values Extended
For certain special angles the trigonometric ratios can be expressed exactly (without decimals), using surds. These must be memorised for the Extended tier exams.
| $\theta$ | $\sin\theta$ | $\cos\theta$ | $\tan\theta$ |
|---|---|---|---|
| $0°$ | $0$ | $1$ | $0$ |
| $30°$ | $\dfrac{1}{2}$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{1}{\sqrt{3}}$ |
| $45°$ | $\dfrac{1}{\sqrt{2}}$ | $\dfrac{1}{\sqrt{2}}$ | $1$ |
| $60°$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{1}{2}$ | $\sqrt{3}$ |
| $90°$ | $1$ | $0$ | undefined |
Memory Trick
For sin: $0°, 30°, 45°, 60°, 90°$ give $\tfrac{\sqrt{0}}{2}, \tfrac{\sqrt{1}}{2}, \tfrac{\sqrt{2}}{2}, \tfrac{\sqrt{3}}{2}, \tfrac{\sqrt{4}}{2}$. The pattern for cos is the reverse. For tan: divide each sin value by the corresponding cos value.
Example 3.6 — Using exact values
A right-angled triangle has hypotenuse $10\,\text{cm}$ and an angle of $60°$. Find the length of the opposite side, leaving your answer in surd form.
Step 1 $\sin 60° = \dfrac{O}{H}$
Step 2 $\dfrac{\sqrt{3}}{2} = \dfrac{O}{10}$
Step 3 $O = 10 \times \dfrac{\sqrt{3}}{2} = \mathbf{5\sqrt{3}\,\text{cm}}$
3.6 Angles of Elevation and Depression
Definitions
- Angle of elevation: the angle measured upward from the horizontal to a line of sight. An observer looking up at a tall building sees the top at an angle of elevation.
- Angle of depression: the angle measured downward from the horizontal to a line of sight. An observer at the top of a cliff looking down at a boat sees it at an angle of depression.
Both angles are always measured from the horizontal. When a diagram has parallel horizontal lines, the angle of depression above equals the angle of elevation below (alternate angles).
Example 3.7 — Angle of elevation
A person stands $30\,\text{m}$ from the base of a vertical tower. The angle of elevation to the top of the tower is $52°$. Find the height of the tower to 3 s.f.
Step 1 Horizontal distance = A = 30, angle = 52°. We need the height = O. Use tan.
Step 2 $\tan 52° = \dfrac{O}{30}$
Step 3 $O = 30 \tan 52° = 30 \times 1.2799\ldots \approx \mathbf{38.4\,\text{m}}$
Example 3.8 — Angle of depression
From the top of a cliff $60\,\text{m}$ high, a boat is observed at an angle of depression of $28°$. How far is the boat from the base of the cliff (measured horizontally)? Give your answer to 3 s.f.
Step 1 The angle of depression from the top equals the angle of elevation from the boat: both are $28°$. Height = O = 60, find horizontal distance = A. Use tan.
Step 2 $\tan 28° = \dfrac{60}{A}$
Step 3 $A = \dfrac{60}{\tan 28°} = \dfrac{60}{0.5317\ldots} \approx \mathbf{113\,\text{m}}$
3D Trigonometry Extended
In three-dimensional problems, the strategy is the same as for 3D Pythagoras: identify two right-angled triangles, solve the first to find an intermediate length or angle, then solve the second to find the final answer.
Two-Triangle Method for 3D Trig
- Identify the 3D solid and the length or angle required.
- Find a right-angled triangle in the base and calculate any needed 2D length (often using Pythagoras).
- Construct a second right-angled triangle that includes the 3D dimension (height of pyramid, space diagonal of cuboid, etc.).
- Apply SOHCAHTOA to the second triangle.
Example 3.9 — Angle in a cuboid
A cuboid has length $8\,\text{cm}$, width $6\,\text{cm}$, and height $5\,\text{cm}$. Find the angle that the space diagonal makes with the base, to 1 d.p.
Step 1 Base diagonal: $d = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\,\text{cm}$
Step 2 The space diagonal, the base diagonal, and the vertical height form a right-angled triangle. The angle $\theta$ at the base satisfies:
$$\tan\theta = \frac{5}{10} = 0.5$$
Step 3 $\theta = \tan^{-1}(0.5) \approx \mathbf{26.6°}$
Exercises
Give all answers to 3 significant figures unless an exact answer is possible or the question specifies otherwise.
Exercise 3.1
In a right-angled triangle, the hypotenuse is $20\,\text{cm}$ and one angle is $37°$. Find the side opposite the $37°$ angle.
Show Solution
O and H → use sin: $\sin 37° = O/20$
$O = 20\sin 37° \approx 20 \times 0.6018 \approx \mathbf{12.0\,\text{cm}}$
Exercise 3.2
A right-angled triangle has an adjacent side of $14\,\text{cm}$ and an angle of $50°$ (between that side and the hypotenuse). Find the length of the hypotenuse.
Show Solution
A = 14, H = ?, A and H → use cos.
$\cos 50° = 14/H \Rightarrow H = 14/\cos 50° \approx 14/0.6428 \approx \mathbf{21.8\,\text{cm}}$
Exercise 3.3
Find angle $\theta$ in a right-angled triangle where the opposite side is $9\,\text{cm}$ and the adjacent side is $12\,\text{cm}$.
Show Solution
O = 9, A = 12 → use tan.
$\tan\theta = 9/12 = 0.75 \Rightarrow \theta = \tan^{-1}(0.75) \approx \mathbf{36.9°}$
Exercise 3.4
A ladder $5\,\text{m}$ long leans against a vertical wall making an angle of $68°$ with the ground. How high up the wall does it reach, to 3 s.f.?
Show Solution
H = 5, angle with ground = 68°. Height = O (opposite the 68° angle).
$\sin 68° = O/5 \Rightarrow O = 5\sin 68° \approx 5 \times 0.9272 \approx \mathbf{4.64\,\text{m}}$
Exercise 3.5
From a point on the ground $50\,\text{m}$ from the base of a vertical flagpole, the angle of elevation to the top is $24°$. Find the height of the flagpole.
Show Solution
A = 50, $\theta = 24°$, find O.
$\tan 24° = O/50 \Rightarrow O = 50\tan 24° \approx 50 \times 0.4452 \approx \mathbf{22.3\,\text{m}}$
Exercise 3.6
A bird sits on a cliff edge $80\,\text{m}$ above the sea. It sees a fish at an angle of depression of $35°$. How far is the fish from the point directly below the bird?
Show Solution
O = 80, $\theta = 35°$ (angle of depression = angle of elevation from the fish).
$\tan 35° = 80/A \Rightarrow A = 80/\tan 35° \approx 80/0.7002 \approx \mathbf{114\,\text{m}}$
Exercise 3.7 Extended
A right-angled triangle has hypotenuse $8\,\text{cm}$ and an angle of $45°$. Find the exact lengths of both shorter sides.
Show Solution
$\sin 45° = O/8 \Rightarrow O = 8 \times \dfrac{1}{\sqrt{2}} = \dfrac{8}{\sqrt{2}} = 4\sqrt{2}\,\text{cm}$
Since $\theta = 45°$, both shorter sides are equal: each has length $\mathbf{4\sqrt{2}\,\text{cm}}$.
Exercise 3.8 Extended
Show that $\cos 30° \times \tan 30° = \sin 30°$, using exact values.
Show Solution
$\cos 30° \times \tan 30° = \dfrac{\sqrt{3}}{2} \times \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{2\sqrt{3}} = \dfrac{1}{2} = \sin 30°$ ✓
Exercise 3.9 Extended
A square-based pyramid has a base of side $10\,\text{cm}$ and a perpendicular height of $12\,\text{cm}$. Find the angle that a slant edge makes with the base, to 1 d.p.
Show Solution
Half the base diagonal: $d = \sqrt{5^2+5^2} = \sqrt{50} = 5\sqrt{2}\,\text{cm}$
The slant edge, half-diagonal, and height form a right-angled triangle.
$\tan\theta = \dfrac{12}{5\sqrt{2}} = \dfrac{12}{7.071\ldots}$
$\theta = \tan^{-1}(1.6971\ldots) \approx \mathbf{59.5°}$
Exercise 3.10 Extended
A cuboid measures $4\,\text{cm}$ by $4\,\text{cm}$ by $7\,\text{cm}$. Find the angle that the space diagonal makes with the longer pair of faces, to 1 d.p.
Show Solution
The base is $4\times4$, so base diagonal $d = \sqrt{4^2+4^2} = 4\sqrt{2}$.
Space diagonal $D = \sqrt{(4\sqrt{2})^2 + 7^2} = \sqrt{32+49} = \sqrt{81} = 9$.
Angle with base: $\tan\theta = 7/(4\sqrt{2}) \Rightarrow \theta = \tan^{-1}(7/(4\sqrt{2})) = \tan^{-1}(1.2374\ldots) \approx \mathbf{51.1°}$
Exam Tips
Tip 1 — Label O, A, H before choosing a ratio
Never select a ratio by guessing. Draw the triangle, mark the given angle $\theta$, then label the three sides O (opposite), A (adjacent), H (hypotenuse). The correct ratio becomes obvious.
Tip 2 — Check calculator is in Degree mode
IGCSE always uses degrees. If your calculator is in Radian mode, $\sin 30°$ will display approximately $-0.988$ instead of $0.5$. Check the mode indicator before every exam.
Tip 3 — Rearranging the formula
Cover up the unknown in the ratio triangle. If you cover O in $\sin\theta = O/H$, you see $H\sin\theta$ — that is, O = H × sin θ. If you cover H, you see O/sin θ — that is, H = O ÷ sin θ. The "formula triangle" works for all three ratios.
Tip 4 — Angles of elevation/depression: look for alternate angles
When a problem shows an observer looking down and a second figure at ground level, the angle of depression equals the angle of elevation (alternate interior angles between parallel horizontal lines). Draw a horizontal line at both levels to make this clear.