Use radians and triangle ratios

Use radians for circular measure

Choose whether the question asks for a length or an area. Both formulas use the same fraction of a full turn, but they scale differently with radius.

The source gives $180^\circ=\pi$ radians. An arc PQ of length s subtends angle θ at the centre O of a circle of radius r. For θ in radians,

$$\frac{\text{arc length}}{\text{circumference}}=\frac\theta{2\pi},\qquad \frac{s}{2\pi r}=\frac\theta{2\pi}\ \Longrightarrow\ s=r\theta.$$

For sector POQ of area A, the corresponding area fraction gives

$$\frac{\text{sector area}}{\text{circle area}}=\frac\theta{2\pi},\qquad \frac A{\pi r^2}=\frac\theta{2\pi}\ \Longrightarrow\ A=\frac12r^2\theta.$$

The two source circle diagrams distinguish the labelled arc length s from the sector area. Both retain O,P,Q,r and the central angle θ. Convert degrees to radians before using either formula.

Match triangle sides to the chosen angle

Mark the angle first, then name its opposite and adjacent sides. Those names depend on the chosen angle; the hypotenuse stays opposite the right angle.

In the source right triangle ABC, B is the right angle and the acute angle θ is at C. Thus AB is opposite, BC adjacent and AC the hypotenuse:

$$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{AB}{AC},\quad \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{BC}{AC},\quad \tan\theta=\frac{\text{opposite}}{\text{adjacent}}=\frac{AB}{BC}=\frac{\sin\theta}{\cos\theta}.$$

The reciprocal ratios are $\operatorname{cosec}\theta=1/\sin\theta$, $\sec\theta=1/\cos\theta$ and $\cot\theta=1/\tan\theta=\cos\theta/\sin\theta$, wherever the corresponding denominators are nonzero. The cotangent ratio cosθ/sinθ also defines cotangent when tanθ itself is undefined but sinθ is nonzero; the reciprocal-of-tangent equality needs both expressions defined.

Use these five exact special-angle cases. Tangent is undefined at π/2 because cosine is zero:

0

sinθ 0

cosθ 1

tanθ 0

$\pi/6$ (30°)

sinθ $1/2$

cosθ $\sqrt3/2$

tanθ $\sqrt3/3$

$\pi/4$ (45°)

sinθ $\sqrt2/2$

cosθ $\sqrt2/2$

tanθ 1

$\pi/3$ (60°)

sinθ $\sqrt3/2$

cosθ $1/2$

tanθ $\sqrt3$

$\pi/2$ (90°)

sinθ 1

cosθ 0

tanθ Undefined

Complementary-angle properties are

$$\sin(\pi/2-\theta)=\cos\theta,\quad \cos(\pi/2-\theta)=\sin\theta,\quad \tan(\pi/2-\theta)=\cot\theta,$$

with the tangent/cotangent statement restricted to defined values.

Use sine and cosine rules in a general triangle

Pair each side with its opposite angle. Use the cosine rule for two sides and their included angle, or three sides. With the sine rule, check whether a second angle gives another valid triangle.

In triangle ABC let lower-case a,b,c label the opposite sides BC, CA, AB respectively. The source sine rule is

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

All three printed cosine-rule forms are

$$\begin{aligned} c^2&=a^2+b^2-2ab\cos C,\\ b^2&=a^2+c^2-2ac\cos B,\\ a^2&=b^2+c^2-2bc\cos A. \end{aligned}$$

The general triangle diagram labels all six angles/sides; it does not require a right angle. Example 16 retains the modified exam attribution and uses the angle at B between sides 1 and 3.