Right-Angled Trigonometry

Right-angled trigonometry deals exclusively with triangles that have one angle of 90°. The right angle fixes the relationship between the three sides, allowing us to calculate unknown lengths and angles from partial information.

The hypotenuse is always the longest side — it lies opposite the right angle. The other two sides are called the adjacent and opposite, and their roles depend on which angle you are focusing on.

Pythagoras' theorem states that in any right-angled triangle, $a^2 + b^2 = c^2$, where $c$ is the hypotenuse. This equation holds regardless of the triangle's size or shape, as long as the right angle is present.

The three trigonometric ratios relate an angle to the ratio of two sides: sin θ = O/H, cos θ = A/H, and tan θ = O/A. Here O = opposite, A = adjacent, H = hypotenuse, all measured relative to angle $\theta$.

The SOHCAHTOA mnemonic helps recall which ratio uses which sides: Sine = Opposite /Hypotenuse, Cosine = Adjacent /Hypotenuse, Tangent = Opposite /Adjacent.

Pythagoras' theorem works because the square on the hypotenuse equals the sum of squares on the other two sides. Geometrically, this expresses a fundamental relationship in Euclidean space that holds for all right-angled triangles.

For finding the hypotenuse: you add inside the square root — $c = \sqrt{a^2 + b^2}$. The hypotenuse is always the longest side, so the result of addition is used.

For finding a shorter side: you subtract inside the square root — $a = \sqrt{c^2 - b^2}$. One leg is the difference between the hypotenuse squared and the other leg squared.

The trig ratios arise from the fact that in similar right-angled triangles, the ratio of any two sides depends only on the angle, not on the triangle's size. This invariance is what makes trigonometry useful.

Inverse functions ($\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$) recover the angle from a known ratio. When you have two sides, you compute the ratio, then apply the inverse to get the angle in degrees or radians.

Finding Missing Lengths

• Step 1: Label the sides O, A, H relative to the given angle $\theta$. The labelling determines which ratio you will use.
• Step 2: Decide which ratio involves the known side and the unknown side. For example, if you know H and want O, use sin: $\sin\theta = \frac{O}{H}$.
• Step 3: Substitute the known values and solve for the unknown. Rearrange the equation to isolate the unknown side.
• Step 4: Verify your answer is reasonable: the hypotenuse must be longest; leg lengths must be positive.

Finding Missing Angles

• Step 1: Label O, A, H for the angle you want to find.
• Step 2: Identify which two sides you know. If you know O and H, use $\theta = \sin^{-1}\left(\frac{O}{H}\right)$.
• Step 3: Use your calculator in degree mode. Ensure you know whether the exam expects degrees or radians.
• Step 4: Check that the angle is in the expected range (0° to 90° for a single acute angle in a right-angled triangle).

Pythagoras vs trig: Use Pythagoras when you have two sides and need the third (and no angle is involved). Use trig when an angle is given or required.

Choosing the ratio: Match the ratio to the sides you have. If both O and H are in play, use sin or sin⁻¹. If A and H, use cos. If O and A, use tan — this avoids needing the hypotenuse.

Exact values are expressions involving $\sqrt{2}$, $\sqrt{3}$, and simple fractions — no decimals. They appear frequently in exam questions and simplify algebraic work.

The 30-60-90 triangle has sides in ratio $1 : \sqrt{3} : 2$. If the shortest side (opposite 30°) has length 1, the side opposite 60° has length $\sqrt{3}$, and the hypotenuse has length 2.

The 45-45-90 triangle has sides in ratio $1 : 1 : \sqrt{2}$. Both legs are equal; the hypotenuse is $\sqrt{2}$ times a leg.

Exact values table (memorise): $\sin 30° = \frac{1}{2}$, $\cos 30° = \frac{\sqrt{3}}{2}$, $\tan 30° = \frac{1}{\sqrt{3}}$; $\sin 45° = \cos 45° = \frac{1}{\sqrt{2}}$, $\tan 45° = 1$; $\sin 60° = \frac{\sqrt{3}}{2}$, $\cos 60° = \frac{1}{2}$, $\tan 60° = \sqrt{3}$.

Derive exact values by applying SOHCAHTOA to the special triangles. For example, in the 30-60-90 triangle with shortest side 1: $\sin 60° = \frac{\sqrt{3}}{2}$ because opposite = $\sqrt{3}$, hypotenuse = 2.

Always start by labelling O, A, H for the angle in question. The same triangle has different O and A depending on which angle you use — a common source of errors.

Check your calculator mode — degrees vs radians. GCSE exams almost always use degrees. If $\sin 30°$ does not give 0.5, you are in the wrong mode.

Sanity check: The hypotenuse must be the longest side. If your "hypotenuse" is shorter than a leg, you have misidentified the sides.

Exact vs decimal: When the question says "give your answer in surd form" or "exact", use $\sqrt{2}$, $\sqrt{3}$, and fractions — never round to decimals.

Inverse trig range: $\sin^{-1}$ and $\cos^{-1}$ return values in specific ranges. For acute angles in right-angled triangles, you will get 0° to 90°, which is correct.

Using the wrong ratio: Students often pick sin when they should use cos, or vice versa. The fix is to always ask: which two sides do I have? Then choose the ratio that contains both.

Hypotenuse confusion: In Pythagoras, mixing up which side is c leads to wrong signs. Remember: c is always the side opposite the right angle.

Subtract in wrong order: For $a = \sqrt{c^2 - b^2}$, you must subtract the smaller square from the larger. If you compute $\sqrt{b^2 - c^2}$ you get an invalid (negative under root) result.

Degrees vs radians: Using radians when degrees are expected (or vice versa) produces completely wrong numerical answers. Always verify the mode before starting.