A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

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.)

  • This Question is Open
  1. zephyr141
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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|

  2. zephyr141
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1444159109427:dw|

  3. zephyr141
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    we're using trig so this is what we got. we have a triangle|dw:1444159195123:dw|

  4. zephyr141
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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\[c=\sqrt{a^2+b^2}\]\[C=\sqrt{(7+15\cos55)^2+\left( \frac{15}{\tan55} \right)^2}\]

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank you. But how would I find the angle? @zephyr141

  6. zephyr141
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |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.

  7. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.