Bong walks 100 m due North and then 150 m in a direction N 37 deg. E. How far is Bong from his original position?

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Answer 1 · 2018-04-19T04:16:09

$color(purple)("Bong is " 181.38 m " away from the original position"$

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Applying Law of Cosines,

$c^2 = a^2 + b^2 - 2 * a * b * cos C$

$" Given : " b = 100 m, a= 150 m, hat C = 180-37 = 143^@$

$c = sqrt((100^2 + 150^2 - (2 * 100 * 150 * cos 143))$

$c = sqrt(10000 + 22500 - 500 cos 143)$

$color(purple)(c = 181.38 m$