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Last updated at April 16, 2024 by Teachoo

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 girlβs displacement from her initial point of departure.Let initial point of departure be O(0, 0) & OA = 4km, AB = 3km β XAB = 30Β° Here, Displacement will be vector (ππ΅) β So, we need to find coordinates of B Rough Representation of 30Β° east of north Now, β BAO = 90Β° β β XAB = 90Β° β 30Β° = 60Β° Let us draw BD β₯ OA In right triangle ADB sin 60Β° = π΅π·/π΄π΅ β3/2 = π΅π·/3 (3β3)/2 = BD BD = (πβπ)/πIn right triangle ADB cos 60Β° = π΄π·/π΄π΅ 1/2 = π΄π·/3 3/2 = AD AD = π/π Now, OD = OA β AD OD = 4 β π/π = (π(π) β π)/π = π/π So, x-coordinate of point B = βOD = (βπ)/π Thus, coordinates of point B = ((βπ)/π, (πβπ)/π) So, Displacement = (βπ)/π π Μ + (πβπ)/π π Μ y-coordinate of point B = BD = (πβπ)/π