1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Easy Trig question that is ridiculously hard!

  1. Aug 18, 2009 #1
    The picture speaks for itself, can anyone explain to me how to do this, they put it near the start of my book and i keep coming back to it yet no sine or cosine formulae etc... can help, please :)
     

    Attached Files:

  2. jcsd
  3. Aug 18, 2009 #2

    berkeman

    User Avatar

    Staff: Mentor

    Start by drawing more lines on an accurate drawing. Draw a line from where the circle intersects the left line, down to where it intersects the middle of the BC line. Look at the triangle that is formed by that line and the angle B. You know B is 50 degrees, you know the two other angles are equal (to what?). Now look at the corresponding angles inside the circle for a triangle from the center of the circle out to those two points -- what are the angles in that circle? Now can you start filling in some of the distances needed?
     
  4. Aug 18, 2009 #3
    (From the book:
    This question comes from chapter 3 and a barely newfound knowledge of sine (soh) and cosine (cah) is expected to explain this. The answer is 8.58cm although i really can't understand how it was accomplished.

    I've attached an attempt in picture form, using this logic i continue and use the sine formula but get a wrong result for BC, idk what to do to get 8.58cm.
     

    Attached Files:

  5. Aug 19, 2009 #4

    berkeman

    User Avatar

    Staff: Mentor

    The angles are labeled wrong. The angles of a triangle add up to 180 degrees. How can one angle be 50 degrees, and the other two angles be 100 degrees apiece?
     
  6. Aug 19, 2009 #5

    berkeman

    User Avatar

    Staff: Mentor

    And your drawing does not show a circle in the middle. Is it a circle?
     
  7. Aug 19, 2009 #6
    Here is another approach. Call the center of the circle O. You know the 2 cm radius of the circle bisects BC. Call that intersection Z. Draw a line from O to C. You now have a triangle OZC. Can you can find the angle OCZ? You know OZ is 2 cm. From the defination of Tangent angle you can find ZC. Twice ZC is BC. It might be helpful to draw a line segment from O to the intersection of the circle and segment AC and indentify all the measures of the interior angles. The answer is indeed 8.58 cm.
     
  8. Aug 19, 2009 #7
    Do you know that the intersection of the symmetric lines of the angles create the center of the circle?

    I do not find this problem hard to solve.

    Here is picture:

    bg7381.png

    just use tan and find BC/2.

    Regards.
     
  9. Aug 19, 2009 #8
    i should have tried it myself, taking tan25'=2/x but i had thought if it and said no to myself, well thanx a lot u've made my day lol :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Easy Trig question that is ridiculously hard!
  1. Hard trig question (Replies: 12)

  2. Easy trig question (Replies: 3)

Loading...