Worked Example 3: Finding Total Distance Using Pythagoras
Part of Bearings — GCSE Mathematics
This study notes covers Worked Example 3: Finding Total Distance Using Pythagoras within Bearings for GCSE Mathematics. Revise Bearings in Geometry & Measures for GCSE Mathematics with 19 exam-style questions and 5 flashcards. This is a high-frequency topic, so it is worth revising until the explanation feels precise and repeatable. It is section 7 of 8 in this topic. Use this study notes to connect the idea to the wider topic before moving on to questions and flashcards.
Topic position
Section 7 of 8
Practice
19 questions
Recall
5 flashcards
Worked Example 3: Finding Total Distance Using Pythagoras
A plane flies on bearing 030° for 100km, then on bearing 120° for 80km. How far is it from its starting point?
Step 1 Find North-South and East-West components for first leg
First leg: bearing 030° for 100km
North component₁ = 100 × cos(30°) = 86.6 km
East component₁ = 100 × sin(30°) = 50 km
Step 2 Find components for second leg
Second leg: bearing 120° for 80km
This is 30° past East, so 60° from South
South component₂ = 80 × cos(60°) = 40 km
East component₂ = 80 × sin(60°) = 69.3 km
Step 3 Add all components
Total North = 86.6 - 40 = 46.6 km
Total East = 50 + 69.3 = 119.3 km
Step 4 Use Pythagoras
Distance² = 46.6² + 119.3² = 2172 + 14232 = 16404
Distance = 128.1 km