Use Pythagoras' theorem in two and three dimensions, use the sine, cosine and tangent ratios to find sides and angles in right-angled triangles, and use exact trigonometric values.

What this dot point is asking

Pythagoras' theorem and right-angled trigonometry find missing lengths and angles in right-angled triangles, a core CCEA Geometry and Measures skill. You must use Pythagoras in two and three dimensions, use the sine, cosine and tangent ratios (SOHCAHTOA) to find both sides and angles, and at Higher tier recall the exact values for the special angles. These tools appear in bearings, mensuration and real contexts, and the three-dimensional cases are reliable Higher-tier marks.

Pythagoras' theorem

In any right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.

The single most important decision is whether you are finding the hypotenuse (add the squares) or a shorter side (subtract). Keeping the hypotenuse identified prevents the most common mistake.

Choosing a trigonometric ratio

When an angle is involved, label the sides relative to that angle: the opposite is across from it, the adjacent is next to it, and the hypotenuse is opposite the right angle.

Finding sides and angles

To find a missing side, choose the ratio that involves the known angle and the known side, then rearrange. To find a missing angle, form the ratio from two known sides and apply the inverse function ($\sin^{-1}$, $\cos^{-1}$ or $\tan^{-1}$).

Three dimensions and exact values (Higher)

In three dimensions, work with a right-angled triangle inside the solid. To find the space diagonal of a cuboid, first find the diagonal of the base by Pythagoras, then use it as one side of a second right-angled triangle with the height. The same idea finds the angle a diagonal makes with the base.

At Higher tier you should know the exact values, such as $\sin 30^\circ = \tfrac{1}{2}$, $\cos 60^\circ = \tfrac{1}{2}$, $\tan 45^\circ = 1$, $\sin 60^\circ = \tfrac{\sqrt{3}}{2}$ and $\cos 30^\circ = \tfrac{\sqrt{3}}{2}$. These appear in non-calculator questions where an exact surd answer is wanted, and they come from two special triangles: the half of an equilateral triangle gives the $30^\circ$ and $60^\circ$ values, and an isosceles right-angled triangle gives $\sin 45^\circ = \cos 45^\circ = \tfrac{1}{\sqrt{2}}$. Closely related to trigonometry are angles of elevation and depression, the angles measured up from or down to the horizontal, which turn a real height-and-distance problem into a right-angled triangle. Sketching that triangle, marking the right angle and the known angle, is the step that makes the choice of ratio obvious.

Why this matters

Pythagoras and trigonometry turn geometry into calculation, underpinning bearings, mensuration, the sine and cosine rules and vector lengths. The three-dimensional and exact-value work links to surds and is a dependable source of Higher-tier marks. CCEA rewards clearly labelled diagrams and a stated ratio, so set out the opposite, adjacent and hypotenuse before calculating.