Pythagoras & SOHCAHTOA
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Objective
Master the geometric foundations of Pythagoras' theorem and right-angled trigonometry (\(\text{SOH CAH TOA}\)) to solve practical distance and angle problems.
Interactive Triangle Lab: Ladder & Wall
Pythagoras: \(a^2 + b^2 = c^2 \implies\) \(6^2 + 8^2 = 100 \implies c = 10.0 \text{ m}\)
Ground Angle (\(\theta\)): \(\tan^{-1}(8/6) = 53.13^\circ\)
Trig Verification: \(\sin(\theta) = \frac{b}{c} =\) \(0.80\)
First Principles & Geometric Intuition
The Golden Right-Triangle Rule: Every right-angled triangle is completely defined by two independent pieces of information (two side lengths, or one side length and one acute angle).
1. Pythagoras' Theorem (Metric Relationship): When you build physical squares along the two shorter legs \(a\) and \(b\) of a right triangle, the total combined area of these two squares exactly fills the square constructed on the longest side (the hypotenuse \(c\)): \[a^2 + b^2 = c^2\]
2. Defining Trigonometric Ratios (Scale-Invariant Angular Ratios): No matter how large or small a right triangle is, if one of its acute angles is \(\theta\), the ratio between pairs of sides remains constant:
- Sine (\(\sin\)): Ratio of the side directly across (Opposite) to the longest side (Hypotenuse): \(\sin\theta = \frac{\text{Opp}}{\text{Hyp}}\).
- Cosine (\(\cos\)): Ratio of the adjacent leg touching the angle to the hypotenuse: \(\cos\theta = \frac{\text{Adj}}{\text{Hyp}}\).
- Tangent (\(\tan\)): Ratio of the steepness/rise (Opposite) over the run (Adjacent): \(\tan\theta = \frac{\text{Opp}}{\text{Adj}}\).
Key Formulas
1. Pythagoras' Theorem
\[c^2 = a^2 + b^2 \iff c = \sqrt{a^2 + b^2}\]\[a = \sqrt{c^2 - b^2}, \quad b = \sqrt{c^2 - a^2}\]Where \(c\) is always the hypotenuse (opposite the \(90^\circ\) angle).
2. Trigonometric Ratios (\(\text{SOH CAH TOA}\))
\[\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \quad (\text{SOH})\]\[\cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} \quad (\text{CAH})\]\[\tan\theta = \frac{\text{Opposite}}{\text{Adjacent}} \quad (\text{TOA})\]3. Inverse Trigonometric Ratios (Finding Unknown Angles)
\[\theta = \sin^{-1}\left(\frac{\text{Opposite}}{\text{Hypotenuse}}\right) = \cos^{-1}\left(\frac{\text{Adjacent}}{\text{Hypotenuse}}\right) = \tan^{-1}\left(\frac{\text{Opposite}}{\text{Adjacent}}\right)\]Ensure calculator is set to DEG (Degrees) mode.
Worked Examples
A builder in Nakuru constructs a rectangular timber roof truss with base leg \(a = 7\text{ m}\) and vertical support \(b = 24\text{ m}\). Find the length of the sloping rafter (hypotenuse \(c\)).
- Identify knowns & unknown: \(a = 7\text{ m}\), \(b = 24\text{ m}\), \(c = ?\).
- Apply Pythagoras' Theorem: \[c^2 = a^2 + b^2\]
- Substitute values: \[c^2 = 7^2 + 24^2 = 49 + 576 = 625\]
- Take square root: \[c = \sqrt{625} = 25\text{ m}\]
A telecommunication transmission mast in Nairobi is supported by a 40-metre wire anchored to the ground. The guy-wire makes an angle of \(60^\circ\) with the horizontal ground. Calculate the vertical height of the mast on the pole to 2 decimal places.
- Identify the sides relative to \(60^\circ\): The mast is Opposite (\(h\)), and the wire is Hypotenuse (\(40\text{ m}\)).
- Select ratio (SOH): \[\sin(60^\circ) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{h}{40}\]
- Rearrange and solve: \[h = 40 \times \sin(60^\circ) = 40 \times \frac{\sqrt{3}}{2} \approx 40 \times 0.866025 = 34.64\text{ m}\]
A ramp for a wheelchair entrance at a clinic in Kisumu rises \(1.5\text{ m}\) vertically over a horizontal run of \(8\text{ m}\).
(a) Find the angle of inclination to the nearest whole degree.
(b) Calculate the total ramp surface length \(L\) to 2 decimal places.
- Part (a) — Find angle \(\theta\) using TOA: \[\tan\theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{1.5}{8} = 0.1875\] \[\theta = \tan^{-1}(0.1875) \approx 10.62^\circ \approx 11^\circ\]
- Part (b) — Find ramp length \(L\) using Pythagoras: \[L = \sqrt{8^2 + 1.5^2} = \sqrt{64 + 2.25} = \sqrt{66.25} \approx 8.14\text{ m}\]
Common Mistakes
Real World
1. Solar Panel Installation (Rooftops in Kenya)
Solar technicians in the Rift Valley position solar panels at an optimal tilt angle (typically \(15^\circ\) to \(20^\circ\)) facing the equator. Using basic trigonometry (\(\tan\theta = \frac{h}{d}\)), installers compute the exact height of mounting brackets needed on flat or pitched roofs.
2. Civil Engineering: Road & Railway Gradients
When engineers survey the Standard Gauge Railway (SGR) cutting through the Great Rift Valley escarpment, gradient limits must strictly not exceed specific percentages or slope angles to prevent train slippage. They use \(\sin\theta = \frac{\Delta h}{\text{track length}}\) to determine safe elevation profiles.
3. Screen Aspect Ratios & Dimensions
Televisions and smartphone screens are advertised by their diagonal dimension (e.g., a 65-inch screen). Using the 16:9 standard ratio and Pythagoras' theorem: \[\text{Width} = 65 \times \frac{16}{\sqrt{16^2 + 9^2}} \approx 56.65\text{ in}, \quad \text{Height} = 65 \times \frac{9}{\sqrt{16^2 + 9^2}} \approx 31.87\text{ in}\]
Practice