Pythagoras and Trigonometry

• Step 1: Write the formula and substitute $c^2 = 7^2 + 8^2$

• Step 2: Square and Add $c^2 = 49 + 64 = 113$

• Step 3: Square Root $c = \sqrt{113} \approx 10.6\text{ m}$ (1 d.p.)

To find a shorter side, use SSS: Square, Subtract, Square root.

Worked Example: Finding a Shorter Side

Calculate the missing side $x$ if the hypotenuse is $13\text{ cm}$ and one shorter side is $5\text{ cm}$.

• Step 1: Write the rearranged formula and substitute $x^2 = 13^2 - 5^2$

• Step 2: Square and Subtract $x^2 = 169 - 25 = 144$

• Step 3: Square Root $x = \sqrt{144} = 12\text{ cm}$

Pythagoras' Theorem in 3D Space

You can fit a remarkably long umbrella inside a small rectangular suitcase if you angle it from the bottom-front-left corner to the top-back-right corner. This longest internal line is called the space diagonal.

To find this diagonal ($d$) in 3D space, you can use the extended formula for a cuboid with dimensions $x, y,$ and $z$:

$$d^2 = x^2 + y^2 + z^2$$

Worked Example: 3D Cuboid Diagonal

A cuboid has dimensions $3\text{ cm} \times 4\text{ cm} \times 6\text{ cm}$. Calculate the space diagonal $d$.

• Step 1: Substitute dimensions into the 3D formula $d^2 = 3^2 + 4^2 + 6^2$

• Step 2: Square and Add $d^2 = 9 + 16 + 36 = 61$

• Step 3: Square Root $d = \sqrt{61} \approx 7.81\text{ cm}$ (3 s.f.)

Leaving the answer as $\sqrt{61}$ means writing it as a surd, providing an exact value without rounding.

Algebraic Pythagoras (Higher Tier)

Higher-tier exam questions frequently hide geometry inside algebra, testing your ability to solve equations alongside spatial reasoning. You must set up Pythagoras' formula, expand the brackets, and solve the resulting quadratic equation.

Worked Example: Algebraic Pythagoras

A right-angled triangle has a hypotenuse $x$ and shorter sides $x-3$ and $x-5$. Calculate $x$.

• Step 1: Set up the equation $x^2 = (x-3)^2 + (x-5)^2$

• Step 2: Expand the brackets $x^2 = (x^2 - 6x + 9) + (x^2 - 10x + 25)$

• Step 3: Simplify to standard quadratic form ($ax^2 + bx + c = 0$) $x^2 = 2x^2 - 16x + 34 \implies x^2 - 16x + 34 = 0$

• Step 4: Solve using the Quadratic Formula $x = \frac{16 \pm \sqrt{(-16)^2 - 4(1)(34)}}{2(1)} = \frac{16 \pm \sqrt{120}}{2}$

• Step 5: Check feasibility $x \approx 13.48$ or $x \approx 2.52$. If $x=2.52$, side $x-3$ would be negative (impossible). Therefore, $x = 13.5$ (3 s.f.).

Trigonometric Ratios (SOH CAH TOA)

How do surveyors measure the height of a mountain without climbing it? They use known distances, measured angles, and basic trigonometry.

In right-angled triangles, the ratios of different side lengths depend entirely on the angles. Before calculating anything, you must correctly label the three sides of the triangle relative to the reference angle $\theta$ using this step-by-step method:

• Step 1: Identify the hypotenuse. This is always the longest side, located directly across from the right angle ($90^\circ$).

• Step 2: Identify the opposite side. This is the side located directly across from the reference angle $\theta$.

• Step 3: Identify the adjacent side. This is the remaining side located next to the reference angle $\theta$ (which is not the hypotenuse).

The acronym SOH CAH TOA helps remember the ratios connecting these sides:

• $\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}$
• $\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}$
• $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$

An angle of elevation is measured upwards from the horizontal, while an angle of depression is measured downwards from the horizontal.

Worked Example: Finding a Missing Side using Angle of Elevation

A kite is flying on a string of length $y$. The angle of elevation from the ground to the kite is $40^\circ$ and it is flying at a vertical height of $8\text{ m}$ (the opposite side). Find the length of the string (the hypotenuse).

• Step 1: Identify the ratio We have the Opposite side ($8\text{ m}$) and want the Hypotenuse ($y$), so use Sine (SOH).

• Step 2: Substitute values $\sin(40^\circ) = \frac{8}{y}$

• Step 3: Rearrange and calculate $y = \frac{8}{\sin(40^\circ)} \approx 12.4\text{ m}$ (3 s.f.)

Finding Missing Angles

To find an unknown angle, you must use inverse trigonometric functions ($\sin^{-1}, \cos^{-1}, \tan^{-1}$) on your calculator.

Worked Example: Finding a Missing Angle

A triangle has an adjacent side of $8\text{ cm}$ and a hypotenuse of $23\text{ cm}$. Find the angle $y$.

• Step 1: Identify the ratio We have Adjacent and Hypotenuse, so use Cosine (CAH).

• Step 2: Substitute values $\cos(y) = \frac{8}{23}$

• Step 3: Use the inverse function $y = \cos^{-1}(\frac{8}{23}) \approx 69.6^\circ$ (1 d.p.)

The Sine Rule (Higher Tier)

Basic SOH CAH TOA only works for right-angled triangles, but the real world is full of non-right-angled triangles. The Sine Rule handles these general triangles whenever you have an opposite pair (a known side and its strictly opposite known angle).

Formula for Finding Sides:

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

Formula for Finding Angles:

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

Worked Example: Finding a Side with the Sine Rule

In triangle $ABC$, Angle $A = 21^\circ$, Angle $B = 35^\circ$, and side $b = 23\text{ cm}$. Find side $a$ (marked $x$).

• Step 1: Set up the ratio using opposite pairs $\frac{x}{\sin 21^\circ} = \frac{23}{\sin 35^\circ}$

• Step 2: Rearrange to isolate $x$ $x = \frac{23}{\sin 35^\circ} \times \sin 21^\circ$

• Step 3: Calculate $x \approx 14.4\text{ cm}$ (3 s.f.)

Worked Example: Finding an Angle with the Sine Rule

In triangle $PQR$, side $p = 6.5\text{ cm}$, side $q = 9.2\text{ cm}$, and Angle $Q = 64^\circ$. Find Angle $P$.

• Step 1: Set up the ratio using the angle formula $\frac{\sin P}{6.5} = \frac{\sin 64^\circ}{9.2}$

• Step 2: Rearrange to isolate $\sin P$ $\sin P = \frac{6.5 \times \sin 64^\circ}{9.2} \approx 0.635...$

• Step 3: Use the inverse sine function $P = \sin^{-1}(0.635...) \approx 39.4^\circ$ (1 d.p.)

Sometimes, finding an angle with the Sine Rule results in the ambiguous case, where two triangles could mathematically exist. If the question specifies the angle is obtuse, calculate $180^\circ - x$ (where $x$ is your calculator's acute answer).

The Cosine Rule (Higher Tier)

What happens when you only know the three sides of a triangle but need an angle? The Sine Rule fails because there is no opposite pair. The Cosine Rule connects all three sides and one angle, working for Side-Angle-Side (SAS) to find a side, or Side-Side-Side (SSS) to find an angle.

Formula:

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

To find an angle, use the rearranged version:

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

Worked Example: Finding a Side (SAS)

Find side $a$ in a triangle with sides $b = 7\text{ cm}$, $c = 9\text{ cm}$, and an included angle $A = 60^\circ$.

• Step 1: Substitute into the formula $a^2 = 7^2 + 9^2 - (2 \times 7 \times 9 \times \cos 60^\circ)$

• Step 2: Simplify $a^2 = 49 + 81 - 63 = 67$

• Step 3: Square Root $a = \sqrt{67} \approx 8.19\text{ cm}$ (3 s.f.)

Worked Example: Finding an Angle (SSS)

Find angle $A$ in a triangle with sides $a = 7.1\text{ km}$, $b = 4.2\text{ km}$, and $c = 3.8\text{ km}$.

• Step 1: Substitute into the rearranged formula $\cos A = \frac{4.2^2 + 3.8^2 - 7.1^2}{2 \times 4.2 \times 3.8}$

• Step 2: Simplify the fraction $\cos A = \frac{17.64 + 14.44 - 50.41}{31.92} = \frac{-18.33}{31.92} \approx -0.574...$

• Step 3: Inverse Cosine $A = \cos^{-1}(-0.574...) \approx 125.0^\circ$ (1 d.p.)

Area of a Triangle (Trigonometric Formula)

Every time you use $\frac{1}{2} \times \text{base} \times \text{height}$, you rely on having a perpendicular height. When you only have angle data, use the trigonometric Area of a Triangle formula. This requires two known sides and the included angle (the angle physically trapped between them).

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

Worked Example: Finding Area

A triangle has sides $9\text{ cm}$ and $6\text{ cm}$ with an included angle of $30^\circ$. Calculate the area.

• Step 1: Substitute into the formula $\text{Area} = 0.5 \times 9 \times 6 \times \sin 30^\circ$

• Step 2: Calculate (using exact value $\sin 30^\circ = 0.5$) $\text{Area} = 27 \times 0.5 = 13.5\text{ cm}^2$

3D Trigonometry

Look at the shadow a slanting pole casts on the ground when the sun is directly overhead. That shadow is its mathematical projection.

In 3D trigonometry, finding the angle between a line and a plane requires identifying the projection of the 3D line onto the 2D plane, extracting a right-angled triangle, and applying standard trigonometry.

Worked Example: Angle between Line and Plane

A cuboid has a rectangular base $ABCD$ measuring $3\text{ cm}$ by $4\text{ cm}$, and a vertical height $CG$ of $6\text{ cm}$. Find the angle between the space diagonal $AG$ and the base $ABCD$.

• Step 1: Identify the projection and calculate its length The projection of $AG$ onto the base is the base diagonal $AC$. Using 2D Pythagoras on the base: $AC = \sqrt{3^2 + 4^2} = \sqrt{25} = 5\text{ cm}$.

• Step 2: Extract the 2D triangle Form the right-angled triangle $ACG$, with base $AC = 5\text{ cm}$ and vertical height $CG = 6\text{ cm}$.

• Step 3: Calculate using SOH CAH TOA We have the Opposite ($6\text{ cm}$) and Adjacent ($5\text{ cm}$) relative to the angle $\theta$ at $A$, so use Tangent (TOA). $\tan(\theta) = \frac{6}{5}$ $\theta = \tan^{-1}(1.2) \approx 50.2^\circ$ (1 d.p.)