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

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

http://www.dummies.com/education/math/trigonometry/laws-of-sines-and-cosines/

Applying Law of Cosines,

$c^{2} = a^{2} + b^{2} - 2 \cdot a \cdot b \cdot cos C$ $c^2 = a^2 + b^2 - 2 * a * b * cos C$

$Given : b = 100 m, a = 150 m, ˆ C = 180 - 37 = 143^{\circ}$ $" Given : " b = 100 m, a= 150 m, hat C = 180-37 = 143^@$

$c = \sqrt{(100^{2} + 150^{2} - (2 \cdot 100 \cdot 150 \cdot cos 143))}$ $c = sqrt((100^2 + 150^2 - (2 * 100 * 150 * cos 143))$

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

$c = 181.38 m$ $color(purple)(c = 181.38 m$