anonymous
  • anonymous
Find the resultant displacement of a bear searching for berries on the mountain. The bear heads 55.0º north of west for 15.0 m; then it turns and heads to the west for another 7.00 m. (Use trigonometry to answer, but remember to draw a diagram to help your understanding.)
Physics
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anonymous
  • anonymous
Find the resultant displacement of a bear searching for berries on the mountain. The bear heads 55.0º north of west for 15.0 m; then it turns and heads to the west for another 7.00 m. (Use trigonometry to answer, but remember to draw a diagram to help your understanding.)
Physics
katieb
  • katieb
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zephyr141
  • zephyr141
first you should draw it out. always draw it out. if you don't fully understand the directions just google them. google is really helpful. i use google everyday when i don't understand something. ok so first things first. draw it.|dw:1444158951593:dw|
zephyr141
  • zephyr141
|dw:1444159109427:dw|
zephyr141
  • zephyr141
we're using trig so this is what we got. we have a triangle|dw:1444159195123:dw|

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zephyr141
  • zephyr141
find x by using trig on the small triangle. remember SOH CAH TOA. we have an angle and a hypotenuse measurement. to find the adjacent side measurement we'll use cosine. \[\cos \theta=\frac{adjacent}{hypotenuse}\]so\[\cos55=\frac{x}{15}\]\[x=15\cos55\]that's the length of the base of the smaller triangle. now we know the that there's 7m to add to find the length of the bigger triangle's base. we can also find the height of the smaller triangle which so happens to be the height of the larger triangle as well. \[\tan55=\frac{15}{y}\]\[y=\frac{15}{\tan55}\]now find the hypotenuse of the larger triangle by using the Pythagorean Theorm.\[a^2+b^2=c^\]or\[c=\sqrt{a^2+b^2}\]\[C=\sqrt{(7+15\cos55)^2+\left( \frac{15}{\tan55} \right)^2}\]
anonymous
  • anonymous
Thank you. But how would I find the angle? @zephyr141
zephyr141
  • zephyr141
|dw:1444357299841:dw|remember SOH CAH TOA? \[Sin \theta=\frac{Opposite}{Hypotenuse}\]\[Cos \theta=\frac{Adjacent}{Hypotenuse}\]\[Tan \theta=\frac{Opposite}{Adjacent}\]you see i capitalized the letters that relate to SOH CAH TOA. now we don't know the angle for this new triangle we have. So in order to find this we can use any of these because we know all the sides of the triangle's lengths. we found the hypotenuse in the last part. just plug them in and solve for \theta. like this,\[\tan\theta=\frac{\left( \frac{15}{\tan55} \right)}{7+15\cos55}\]\[\theta=\tan^{-1} \left( \frac{ \left( \frac{ 15 }{ \tan55 } \right) }{ 7+15\cos55 } \right)\]\[\theta=\tan^{-1}\left( \frac{10.5}{15.6} \right)=\tan^{-1}(0.67)=33.9\]so our angle that we find is 33.9 degrees.

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