By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
Mastering Pythagoras and SOH CAH TOA unlocks 10-15% of your GCSE Maths exam marks—and real-life problems like measuring roofs, navigation, or even gaming physics. One question could be the difference between a Grade 5 and a Grade 7.
Formula: a² + b² = c² - a and b = shorter sides (legs). - c = hypotenuse (longest side). MEMORISE THIS – Not given on exam sheets.
a² + b² = c²
a
b
c
Formulas: - Sine (SOH): sin(θ) = Opposite / Hypotenuse - Cosine (CAH): cos(θ) = Adjacent / Hypotenuse - Tangent (TOA): tan(θ) = Opposite / Adjacent MEMORISE THIS – Not given on exam sheets.
sin(θ) = Opposite / Hypotenuse
cos(θ) = Adjacent / Hypotenuse
tan(θ) = Opposite / Adjacent
Formula: d = √(a² + b² + c²) - d = space diagonal (longest diagonal in a cuboid). - a, b, c = length, width, height. MEMORISE THIS – Sometimes given, but not always.
d = √(a² + b² + c²)
d
a, b, c
θ
sin⁻¹
√(a² + b²)
Question: A right-angled triangle has sides 3 cm and 4 cm. Find the hypotenuse.
Solution: 1. Label sides: a = 3, b = 4, c = ?. 2. Write formula: 3² + 4² = c². 3. Substitute: 9 + 16 = c². 4. Add: 25 = c². 5. Square root: c = √25 = 5 cm.
a = 3
b = 4
c = ?
3² + 4² = c²
9 + 16 = c²
25 = c²
c = √25 = 5 cm
What we did and why: We used Pythagoras because it’s a right-angled triangle. We squared the two shorter sides, added them, and took the square root to find the hypotenuse.
Question: In a right-angled triangle, the hypotenuse is 10 cm, and the angle is 30°. Find the opposite side.
Solution: 1. Label sides: Opposite = ?, Hypotenuse = 10, Angle = 30°. 2. Choose ratio: SOH (sin(θ) = Opposite / Hypotenuse). 3. Write formula: sin(30°) = Opposite / 10. 4. Substitute: 0.5 = Opposite / 10 (since sin(30°) = 0.5). 5. Rearrange: Opposite = 10 × 0.5 = 5 cm.
sin(30°) = Opposite / 10
0.5 = Opposite / 10
sin(30°) = 0.5
Opposite = 10 × 0.5 = 5 cm
What we did and why: We used SOH because we had the hypotenuse and needed the opposite side. We substituted the known values and solved for the unknown.
Question: A cuboid has length 3 cm, width 4 cm, and height 12 cm. Find the space diagonal.
Solution: 1. Draw the cuboid and label sides: a = 3, b = 4, c = 12. 2. Find the 2D diagonal of the base: √(3² + 4²) = √(9 + 16) = √25 = 5 cm. 3. Use the 2D diagonal as one side: d = √(5² + 12²). 4. Calculate: d = √(25 + 144) = √169 = 13 cm.
c = 12
√(3² + 4²) = √(9 + 16) = √25 = 5 cm
d = √(5² + 12²)
d = √(25 + 144) = √169 = 13 cm
What we did and why: We used 3D Pythagoras because we needed the longest diagonal in a 3D shape. We first found the 2D diagonal, then used it with the height to find the space diagonal.
c² = 25
c = 5
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