Introduction
This O-Level A-Maths question tests your understanding of:
- the \(\tan(A + B)\) identity
- exact trigonometric values
- using right-angled triangle relationships
You are required to work without a calculator, so accuracy and correct identities are crucial.
We follow the teacher’s exact board working step by step.
The Question
Two acute angles \(A\) and \(B\) are such that:
\(\tan(A + B) = 8,\quad \tan B = 3\)
Without using a calculator, find the exact value of \(\sin A\).
Step-by-Step Working (Teacher’s Method)
Step 1: Use the \(\tan(A + B)\) identity
\(\tan(A + B) = \dfrac{\tan A + \tan B}{1 – \tan A \cdot \tan B}\)
Substitute the given values:
\(8 = \dfrac{\tan A + 3}{1 – 3\tan A}\)
Step 2: Cross-multiply
\(8(1 – 3\tan A) = \tan A + 3\)
\(8 – 24\tan A = \tan A + 3\)
Step 3: Solve for \(\tan A\)
\(8 – 3 = \tan A + 24\tan A\)
\(5 = 25\tan A \quad \Rightarrow \quad \tan A = \tfrac{1}{5}\)
Step 4: Draw a right-angled triangle
For angle \(A\):
\(\tan A = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{1}{5}\)
- Opposite = 1
- Adjacent = 5
Hypotenuse:
\(\sqrt{1^2 + 5^2} = \sqrt{26}\)
Step 5: Find \(\sin A\)
\(\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{1}{\sqrt{26}}\)
✅ Final Answer
\(\sin A = \dfrac{1}{\sqrt{26}}\)
Key Concepts
| Concept | Why It Matters |
|---|---|
| \(\tan(A+B)\) identity | Relates \(\tan A\) and \(\tan B\) |
| Exact values | Calculator not allowed |
| Triangle construction | Converts ratios into trig values |
| Pythagoras’ theorem | Finds the hypotenuse |
Tips for Students
- Memorise the \(\tan(A+B)\) formula
- Always cross-multiply carefully
- Draw a triangle after finding a trig ratio
- Leave answers in exact form
For Parents
This type of question strengthens:
- algebraic manipulation
- trigonometric reasoning
- exam confidence
It is a core skill frequently tested in O-Level A-Maths.