Trigonometry and Pythagoras in 2D and 3D

• This 3D Pythagoras formula is not provided on the Edexcel formula sheet; you must memorise it or derive it.
• Alternatively, use a two-step method by identifying two 2D right-angled triangles:
  1. Apply Pythagoras to the base to find the diagonal of the base ($x^2 = l^2 + w^2$).
  2. Use that base diagonal and the vertical height ($h$) to find the space diagonal ($d^2 = x^2 + h^2$).
• In a square-based pyramid, a right-angled triangle is formed by the perpendicular height, half the base diagonal, and the slant edge (the distance from the apex to a corner of the base).

The Sine Rule

• The Sine Rule is used to find missing sides and angles in a general triangle (a triangle that does not necessarily contain a right angle).
• It is applied when you know an opposite pair (an angle and the side directly across from it) plus one other side or angle.
• To find a missing side, put the unknown side on top of the fraction:

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

• To find a missing angle, put the unknown angle on top:

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

• Finding an angle using the Sine Rule can create an ambiguous case where two different triangles can be formed (one acute, one obtuse). If the diagram shows an obtuse angle but your calculator gives an acute angle $x$, subtract it from $180^\circ$ because $\sin x = \sin(180 - x)$.

The Cosine Rule

• The Cosine Rule provides a unique solution for any angle between $0^\circ$ and $180^\circ$, so it does not have an ambiguous case.
• Use it to find a missing side when given two sides and the included angle (the angle trapped between two known side lengths):

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

• Use it to find a missing angle when given all three sides. You can rearrange the formula to:

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

• Both the Sine Rule and the Cosine Rule are provided on the Edexcel Higher Tier formula sheet.

3D Trigonometry (Right-Angled)

• To find the angle between a line and a plane, you must find the angle between the line and its orthogonal projection (its "shadow") onto that plane.
• This angle $\theta$ is mathematically related to the normal (a line perpendicular to the plane) by $\theta = 90^\circ - \phi$, where $\phi$ is the angle between the line and the normal.
• General Method:
  1. Drop a perpendicular from the top vertex to the plane to form a right-angled triangle.
  2. Use Pythagoras to find the base (projection) length.
  3. Use SOH CAH TOA (usually Tan) to find the required angle.

Applying Sine and Cosine Rules in 3D

• In higher-tier problems, a 3D shape may contain a general triangle that is not right-angled.
• You will often need to use Pythagoras' Theorem on the right-angled faces first to find the side lengths of the general triangle hidden inside the 3D space.
• Once you have the necessary sides or angles, apply the Sine or Cosine Rule to the internal general triangle.

Worked Example: Calculating an Angle in a 3D Shape

The diagram shows a tetrahedron $ABCD$. $AD$ is perpendicular to $AB$ and $AC$. Angle $BAC = 90^\circ$. The side lengths are $AB = 10\text{ cm}$, $AC = 8\text{ cm}$, and $AD = 5\text{ cm}$. Calculate the size of angle $BDC$.

Step 1: Find the three sides of $\triangle BDC$ using Pythagoras' Theorem on the right-angled faces. Keep values as exact surds.

• In $\triangle ABD$ (right-angled at $A$): $BD^2 = 10^2 + 5^2 = 125 \implies BD = \sqrt{125}\text{ cm}$
• In $\triangle ACD$ (right-angled at $A$): $CD^2 = 8^2 + 5^2 = 89 \implies CD = \sqrt{89}\text{ cm}$
• In $\triangle ABC$ (the base, right-angled at $A$): $BC^2 = 10^2 + 8^2 = 164 \implies BC = \sqrt{164}\text{ cm}$

Step 2: Apply the Cosine Rule to the general triangle $\triangle BDC$, as we now have all three sides (SSS).

$$\cos(\angle BDC) = \frac{b^2 + c^2 - a^2}{2bc}$$

Step 3: Substitute the exact surd values. The side opposite the target angle $\angle BDC$ is $BC$ ($a = \sqrt{164}$).

• $\cos(\angle BDC) = \frac{(\sqrt{125})^2 + (\sqrt{89})^2 - (\sqrt{164})^2}{2(\sqrt{125})(\sqrt{89})}$
• $\cos(\angle BDC) = \frac{125 + 89 - 164}{2(\sqrt{125})(\sqrt{89})}$
• $\cos(\angle BDC) = \frac{50}{210.95...} \approx 0.237...$

Step 4: Calculate the inverse cosine to find the angle.

• $\angle BDC = \cos^{-1}(0.237...)$
• $\angle BDC \approx 76.3^\circ$ (to 1 d.p.)