Pythagoras and Trigonometry

In a right-angled triangle with legs $a$, $b$ and hypotenuse $c$:

$a^2+b^2=c^2$

The hypotenuse is the longest side, opposite the right angle.

Finding the hypotenuse: $c=\sqrt{a^2+b^2}$

Worked example: $a=5$, $b=12$. $c=\sqrt{25+144}=\sqrt{169}=13$ cm ✓

Finding a shorter side: $a=\sqrt{c^2-b^2}$

Worked example: $c=10$, $b=6$. $a=\sqrt{100-36}=\sqrt{64}=8$ cm ✓

3D Pythagoras (Higher): apply the theorem twice — first in a horizontal or vertical plane, then use that result as a side in a second triangle.

Worked example — find the length of the space diagonal of a cuboid $12\times 4\times 3$ cm.

Base diagonal: $\sqrt{12^2+4^2}=\sqrt{160}$. Space diagonal: $\sqrt{160+9}=\sqrt{169}=13$ cm ✓

For an acute angle $\theta$ in a right-angled triangle:

$\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}\tan \theta =\frac{\text{opposite}}{\text{adjacent}}$

Finding a side: multiply or divide as appropriate.

Worked example — in a right-angled triangle, $\theta =35°$, hypotenuse $=12$ cm. Find the side opposite $\theta$.

$\sin 35°=\frac{\text{opp}}{12}⟹\text{opp}=12\sin 35°=12\times 0.5736\approx 6.88$ cm ✓

Finding an angle: use the inverse trig function.

Worked example — opposite $=7$ cm, adjacent $=10$ cm. Find $\theta$.

$\tan \theta =7/10=0.7⟹\theta ={\tan }^{-1}(0.7)\approx 35.0°$ ✓

Setting up: correctly identify which sides are opposite, adjacent, and hypotenuse relative to the angle you are working with — not relative to some other angle in the triangle.

These exact values must be memorised — they are not given on the exam formula sheet:

$\theta$	$\sin \theta$	$\cos \theta$	$\tan \theta$
$0°$	$0$	$1$	$0$
$30°$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$
$45°$	$\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$	$\frac{\sqrt{2}}{2}$	$1$
$60°$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$
$90°$	$1$	$0$	undefined

Memory aid: for $\sin$, the values $0,\frac{1}{2},\frac{\sqrt{2}}{2},\frac{\sqrt{3}}{2},1$ come from $\frac{\sqrt{0}}{2},\frac{\sqrt{1}}{2},\frac{\sqrt{2}}{2},\frac{\sqrt{3}}{2},\frac{\sqrt{4}}{2}$.

Worked example — find the exact area of an isosceles right-angled triangle with equal sides 6 cm.

$A=\frac{1}{2}\times 6\times 6=18$ cm². Hypotenuse $=6\sqrt{2}$ cm. ✓

Use non-right-angled triangle rules when no right angle is present.

Sine rule (use when given two angles + one side, or two sides + an angle not between them):

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

Worked example — in triangle $ABC$: $A=48°$, $B=67°$, $a=9$ cm. Find $b$.

$C=180°-48°-67°=65°$. $\frac{b}{\sin 67°}=\frac{9}{\sin 48°}⟹b=\frac{9\sin 67°}{\sin 48°}=\frac{9\times 0.9205}{0.7431}\approx 11.14$ cm ✓

Cosine rule (use when given two sides + the included angle, or all three sides to find an angle):

$a^2=b^2+c^2-2bc\cos A$

Rearranged to find angle: $\cos A=\frac{b^2+c^2-a^2}{2bc}$

Worked example — $b=7$, $c=5$, $A=110°$. Find $a$.

$a^2=49+25-2(7)(5)\cos 110°=74-70(-0.3420)=74+23.94=97.94⟹a\approx 9.90$ cm ✓

For any triangle with two sides $a$ and $b$ and the included angle $C$:

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

This formula applies to any triangle — not just right-angled ones.

Worked example — triangle with sides 8 cm and 11 cm and included angle $42°$:

$A=\frac{1}{2}(8)(11)\sin 42°=44\times 0.6691\approx 29.44$ cm² ✓

Worked example — find angle $C$ if the area is 30 cm², and $a=6$, $b=12$:

$30=\frac{1}{2}(6)(12)\sin C=36\sin C⟹\sin C=\frac{30}{36}=\frac{5}{6}⟹C={\sin }^{-1}(5/6)\approx 56.4°$ ✓

1. Opposite and adjacent — wrong identification

The "opposite" and "adjacent" labels are relative to the angle $\theta$ in question, not to any fixed position. Redraw the triangle and label $\theta$ before assigning sides.

2. Sine rule — ambiguous case

The sine rule can give two possible triangles (acute and obtuse angles with the same sine). If two sides and a non-included angle are given and $\sin A>\sin B$ but $a<b$, check whether $180°-\text{(found angle)}$ also makes sense.

3. Cosine rule — sign error on the final term

$a^2=b^2+c^2-2bc\cos A$: when $A>90°$, $\cos A$ is negative, so the term $-2bc\cos A$ becomes positive. Missing this makes the triangle shorter than it should be.

4. Exact values — confusing sin and cos for 30° and 60°

$\sin 30°=\frac{1}{2}$, $\cos 30°=\frac{\sqrt{3}}{2}$; these swap for 60°. A quick check: $\sin 60°>\sin 30°$ because 60° is closer to 90° (the maximum).

Mistake	Correction
"$\sin 45°=\frac{1}{2}$"	$\sin 45°=\frac{\sqrt{2}}{2}\approx 0.707$; $\sin 30°=\frac{1}{2}$
"Cosine rule: $a^2=b^2-c^2+2bc\cos A$"	$a^2=b^2+c^2-2bc\cos A$ (both $b^2$ and $c^2$ are added)
"Area = $\frac{1}{2}bh$ for any triangle, using a slant side as $h$"	$h$ must be the perpendicular height; use $\frac{1}{2}ab\sin C$ when the perpendicular height is unknown