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Misc 3 - A girl walks 4 km west, then 3 km 30 east of north

Misc 3 - Chapter 10 Class 12 Vector Algebra - Part 2
Misc 3 - Chapter 10 Class 12 Vector Algebra - Part 3

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Misc 3 A girl walks 4 km towards west, then she walks 3 km in a direction 30Β° east of north and stops. Determine the girl’s displacement from her initial point of departure.Let initial point of departure be O(0, 0) & OA = 4km, AB = 3km ∠ XAB = 30Β° Here, Displacement will be vector (𝑂𝐡) βƒ— So, we need to find coordinates of B Rough Representation of 30Β° east of north Now, ∠ BAO = 90Β° – ∠ XAB = 90Β° – 30Β° = 60Β° Let us draw BD βŠ₯ OA In right triangle ADB sin 60Β° = 𝐡𝐷/𝐴𝐡 √3/2 = 𝐡𝐷/3 (3√3)/2 = BD BD = (πŸ‘βˆšπŸ‘)/𝟐In right triangle ADB cos 60Β° = 𝐴𝐷/𝐴𝐡 1/2 = 𝐴𝐷/3 3/2 = AD AD = πŸ‘/𝟐 Now, OD = OA – AD OD = 4 – πŸ‘/𝟐 = (πŸ’(𝟐) βˆ’ πŸ‘)/𝟐 = πŸ“/𝟐 So, x-coordinate of point B = βˆ’OD = (βˆ’πŸ“)/𝟐 Thus, coordinates of point B = ((βˆ’πŸ“)/𝟐, (πŸ‘βˆšπŸ‘)/𝟐) So, Displacement = (βˆ’πŸ“)/𝟐 π’Š Μ‚ + (πŸ‘βˆšπŸ‘)/𝟐 𝒋 Μ‚ y-coordinate of point B = BD = (πŸ‘βˆšπŸ‘)/𝟐

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