At GCSE, $\sin$, $\cos$ and $\tan$ were ratios of sides in a right-angled triangle. At Level 3 they become functions defined for every angle, with graphs, identities and equations of their own.
Radians are not required. Check your calculator is in DEG mode.
| Word | Meaning |
|---|---|
| Hypotenuse | The longest side of a right-angled triangle, opposite the right angle. |
| Opposite / adjacent | Defined relative to the angle you are using. |
| Identity | A relationship true for every angle; written with $\equiv$. |
| Period | The horizontal length of one complete cycle of a graph. |
| Amplitude | Half the vertical distance between the maximum and minimum. |
| Ambiguous case | When the sine rule gives two valid triangles. |
| Angle of elevation | The angle up from the horizontal. |
| Angle of depression | The angle down from the horizontal. |
| Angle of greatest slope | The steepest line of ascent on a sloping plane. |
Pythagoras: $a^2 + b^2 = c^2$
Two sides and the angle between them, or all three sides? → cosine rule
A matching side-and-angle pair? → sine rule
In triangle $ABC$, $AB = 7$ cm, $BC = 9$ cm and angle $B = 62°$. Find $AC$ and then angle $A$.
- PT1Ratios of Any Angle$\sin$, $\cos$ and $\tan$ beyond $90°$, and their graphs.
- PT2The Sine and Cosine RulesNon-right-angled triangles, including the ambiguous case.
- PT3
PT4Trigonometric Identities$\tan\theta \equiv \tfrac{\sin\theta}{\cos\theta}$ and $\sin^2\theta + \cos^2\theta \equiv 1$. - PT5Trigonometric EquationsFinding every solution in a given interval.
- PT62-D and 3-D ProblemsModelling with Pythagoras and trigonometry, including greatest slope.
Degrees
Always. Check DEG mode.
Sine rule
Needs a matching side–angle pair.
Cosine rule
Two sides and the included angle, or all three sides.
Area
$\tfrac12 ab\sin C$, with $C$ between $a$ and $b$.
Identity 1
$\tan\theta \equiv \tfrac{\sin\theta}{\cos\theta}$.
Identity 2
$\sin^2\theta + \cos^2\theta \equiv 1$.
Periods
$\sin$, $\cos$: $360°$. $\tan$: $180°$.
Second solution
Sine: $180° - \theta$. Cosine: $360° - \theta$.
Ambiguous case
Check $180° - \theta$ still fits the triangle.
3-D problems
Redraw the relevant triangle flat, on its own.
A right-angled triangle has legs $5$ cm and $12$ cm. Find the hypotenuse.
▶ Show solution
$c^2 = 25 + 144 = 169$, so $c = 13$ cm.
Find $x$ in a right-angled triangle where the hypotenuse is $10$ cm and the angle between it and $x$ is $35°$.
▶ Show solution
$x$ is adjacent to the $35°$ angle, so $\cos 35° = \dfrac{x}{10}$.
$x = 10\cos 35° = 8.19$ cm (3 s.f.)
In triangle $ABC$, $a = 8$, $A = 50°$ and $B = 70°$. Find $b$.
▶ Show solution
Sine rule: $\dfrac{b}{\sin 70°} = \dfrac{8}{\sin 50°}$
$b = \dfrac{8 \sin 70°}{\sin 50°} = \dfrac{8 \times 0.93969}{0.76604} = 9.81$ (3 s.f.)
A triangle has sides $6$, $7$ and $9$. Find its largest angle.
▶ Show solution
The largest angle is opposite the longest side, $9$.
$\cos\theta = \dfrac{36 + 49 - 81}{2(6)(7)} = \dfrac{4}{84} = 0.047619$
$\theta = 87.3°$ (1 d.p.)
Find the area of a triangle with sides $8$ cm and $11$ cm enclosing an angle of $40°$.
▶ Show solution
Area $= \tfrac12(8)(11)\sin 40° = 44 \times 0.64279$
$= 28.3$ cm² (3 s.f.)
Given $\sin\theta = 0.6$ and $\theta$ is acute, find $\cos\theta$ exactly.
▶ Show solution
$\cos^2\theta = 1 - 0.36 = 0.64$
$\cos\theta = 0.8$ (positive, since $\theta$ is acute).
Solve $\sin x = 0.5$ for $0° \leqslant x \leqslant 360°$.
▶ Show solution
$\sin^{-1}(0.5) = 30°$
The second solution is $180° - 30° = 150°$.
$x = 30°$ or $150°$.
State the period and amplitude of $y = 5\sin(2x)$.
▶ Show solution
Amplitude $5$; period $\dfrac{360°}{2} = 180°$.
The specification's own example: in triangle $ABC$, $AB = 10$ m, $AC = 8$ m and angle $B = 40°$. Find the two possible values of angle $C$.
▶ Show solution
We know side $AC = 8$ is opposite angle $B = 40°$, and side $AB = 10$ is opposite angle $C$.
Sine rule: $\dfrac{\sin C}{10} = \dfrac{\sin 40°}{8}$
$\sin C = \dfrac{10 \times 0.64279}{8} = 0.80349$
$C = 53.5°$ or $C = 180° - 53.5° = 126.5°$ (1 d.p.)
Both are valid: adding $40°$ to either still leaves a positive third angle ($86.5°$ or $13.5°$), so two different triangles fit the data. This is the ambiguous case.
A vertical mast $BT$ stands on horizontal ground. From a point $A$ on the ground, $50$ m from the base $B$, the angle of elevation of the top $T$ is $32°$.
(a) Find the height of the mast. (b) A second point $C$ is $80$ m from $B$, with angle $ABC = 90°$. Find the distance $AC$. (c) Find the angle of elevation of $T$ from $C$. (d) Find the angle that $AT$ makes with the ground at $A$ — and explain why this is not the same as the angle $TAC$.
▶ Show solution
(a) In the right-angled triangle $ABT$: $\tan 32° = \dfrac{BT}{50}$
$BT = 50\tan 32° = 50 \times 0.62487 = 31.24$ m, so the mast is $31.2$ m tall (3 s.f.).
(b) Triangle $ABC$ is right-angled at $B$, with $AB = 50$ and $BC = 80$:
$AC = \sqrt{50^2 + 80^2} = \sqrt{2500 + 6400} = \sqrt{8900} = 94.34$ m (2 d.p.)
(c) In the right-angled triangle $CBT$, with $CB = 80$ and $BT = 31.24$:
$\tan(\text{elevation}) = \dfrac{31.24}{80} = 0.39053$
Elevation $= 21.3°$ (1 d.p.)
(d) The angle $AT$ makes with the ground, measured in the vertical plane through $A$ and $B$, is the angle $TAB = 32°$ given in the question.
The angle $TAC$ is a different angle: it is measured in the plane containing $A$, $T$ and $C$, and $AC$ is a line along the ground in a different direction from $AB$.
To see the difference, note $AT = \dfrac{50}{\cos 32°} = 58.96$ m, while $CT = \sqrt{80^2 + 31.24^2} = 85.88$ m. In triangle $ATC$ we then have all three sides, so use the cosine rule:
$\cos(TAC) = \dfrac{AT^2 + AC^2 - CT^2}{2 \times AT \times AC} = \dfrac{3476 + 8900 - 7376}{2 \times 58.96 \times 94.34} = 0.4495$
$\angle TAC = 63.3°$ — very different from $32°$.
The lesson: in a 3-D problem the "angle of elevation" is always measured in the vertical plane containing the observer and the base of the object. An angle to a point in another direction is a genuinely different angle, and you must identify the correct triangle before calculating.