QUESTION

CSAT

Hard

Maths

Prelims 2016

A person climbs a hill in a straight path from point ‘O’ on the ground in the direction of north-east and reaches a point ‘A’ after travelling a distance of 55 km. Then, from the point ‘A’ he moves to point ‘B’ in the direction of north-west. Let the distance AB be 1212 km. Now, how far is the persons away from the starting point ‘O’?

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Explanation

Distance from O to A = 5 km Distance from A to B = 12 km Angle between north-east and north-west = 90° (which implies cos(90°)=0\cos(90°) = 0) Formula:

Use the Law of Cosines formula: c2=a2+b22abcos(θ)c^2 = a^2 + b^2 - 2ab \cdot \cos(\theta) Where: a=5a = 5 b=12b = 12 θ=90\theta = 90

Substitute Values: c2=52+1222512cos(90°)c^2 = 5^2 + 12^2 - 2 \cdot 5 \cdot 12 \cdot \cos(90°) c2=25+1440c^2 = 25 + 144 - 0 c2=169c^2 = 169 c=169=13c = \sqrt{169} = 13

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