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lala2

  • 2 years ago

help?!?

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  1. phi
    • 2 years ago
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    what is angle B ?

  2. phi
    • 2 years ago
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    angle B= 180 - 24 can you find angle ACB knowing A is 16, and B is 156 ?

  3. phi
    • 2 years ago
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    remember 3 angles of a triangle add up to 180º

  4. phi
    • 2 years ago
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    if you can find angle ACB, you can use the Law of Sines to find the side BC \[ \frac{BC}{\sin 16}= \frac{7600}{\sin( ACB)} \]

  5. phi
    • 2 years ago
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    can you be more specific ?

  6. phi
    • 2 years ago
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    yes, angle at C is 8º now use the Law of Sines to find the length of side BC

  7. lala2
    • 2 years ago
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    bc/sin16= 7600/ sin (16)(156)(8)

  8. phi
    • 2 years ago
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    **im not typing i dont know why its saying that** that's scary. what is this bc/sin16= 7600/ sin (16)(156)(8) ?? \[ \frac{BC}{\sin 16}= \frac{7600}{\sin( 8º)} \]

  9. lala2
    • 2 years ago
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    hah! iknow

  10. phi
    • 2 years ago
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    no, I was trying to say the angle at C (start at A go to C, then go to B) the Law of Sines see http://www.mathsisfun.com/algebra/trig-sine-law.html uses the ratio of the sin(angle) / side opposite in this case, we want to know side BC, so we need to know the angle across from BC. we do , it is 16º we know side AB= 7600, so we want to know the angle opposite that side. it is angle C=8

  11. phi
    • 2 years ago
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    if you work it through, you can write \[ \frac{BC}{\sin (16º)}= \frac{7600}{\sin( 8º)} \] multiply both sides by sin(16). \[ \frac{BC}{\sin (16º) }\cdot \sin(16º)= \frac{7600}{\sin( 8º)} \cdot \sin(16º)\]

  12. phi
    • 2 years ago
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    on the left side, sin(16)/sin(16) "cancel" (become 1 so we can ignore) now get a calculator and figure out BC

  13. phi
    • 2 years ago
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    what were those 2's they are little º meaning "degrees"

  14. phi
    • 2 years ago
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    no. start with \[ \frac{BC}{\sin (16º) }\cdot \sin(16º)= \frac{7600}{\sin( 8º)} \cdot \sin(16º) \]

  15. phi
    • 2 years ago
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    only on the left side you should remember that anything divided by itself is 1 so the left side \[ \frac{BC}{\sin (16º) }\cdot \sin(16º) = BC \cdot \frac{\sin(16)}{\sin(16)}= BC \cdot 1 = BC \]

  16. phi
    • 2 years ago
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    you need a calculator to figure out the right side

  17. phi
    • 2 years ago
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    yes, but the sin(16) is up top 7600*sin(16)/sin(8)

  18. phi
    • 2 years ago
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    yes. but I think they want to the nearest foot

  19. phi
    • 2 years ago
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    you can use the Law of Sines to do part (b)

  20. phi
    • 2 years ago
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    in triangle BCD , what side do you know ?

  21. phi
    • 2 years ago
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    are you looking at the same picture I am ? put you finger on B, go to C, down to D, over to B. that triangle. which of its 3 sides do you know ?

  22. phi
    • 2 years ago
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    yes, but what did you figure out in part (a) ?

  23. phi
    • 2 years ago
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    You found the length of side BC in part (a) BC= 15052.07

  24. lala2
    • 2 years ago
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    oh!

  25. phi
    • 2 years ago
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    do you know the angle that is opposite to side BC ?

  26. lala2
    • 2 years ago
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    no.. how do i find that?

  27. phi
    • 2 years ago
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    the angle opposite side BC is the angle that is not B and not C

  28. lala2
    • 2 years ago
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    im sorry but i just dont understand..

  29. phi
    • 2 years ago
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    what is the *angle* opposite side BC ? you have 3 choices: B , C or D. (and it's not B or C)

  30. phi
    • 2 years ago
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    |dw:1369784436661:dw|

  31. lala2
    • 2 years ago
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    i thought you meant side not angle

  32. phi
    • 2 years ago
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    that means we can write down \[ \frac{15052.07}{\sin(90)} = \] now you need a side (I would pick the one we are looking for, and its opposite angle)

  33. lala2
    • 2 years ago
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    right?

  34. phi
    • 2 years ago
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    yes, write CD over sin(24) and set that equal to the first ratio (up above) what do you get ?

  35. phi
    • 2 years ago
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    I don't know what that is. You should be writing down the equation that you get when you use the Law of Sines. In part (a) we used \[ \frac{BC}{\sin (16)}= \frac{7600}{\sin( 8)} \] your job is to write a similar equation for part (b) you already have both sides....

  36. phi
    • 2 years ago
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    *both sides of the equation \[ \frac{15052.07}{\sin(90)} = \] and ....

  37. phi
    • 2 years ago
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    yes, but use that in the equation... CD/sin(24) is part of the equation

  38. phi
    • 2 years ago
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    the idea is length/sin(angle) = length/ sin(angle)

  39. phi
    • 2 years ago
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    perfect. now "solve" for CD multiply both sides by sin(24) (this will "cancel" sin(24) on the right side)

  40. phi
    • 2 years ago
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    that is the right side of the equation. you should also show the other side. in other words, start with \[ \frac{15052.07}{\sin(90)} = \frac{CD}{\sin(24)} \] multiply both sides by sin(24). the right side will simplify to CD, just like you showed what do you get for the left side of the equation ?

  41. phi
    • 2 years ago
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    not after you multiply it by sin(24)

  42. phi
    • 2 years ago
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    \[ \frac{15052.07}{\sin(90)} = \frac{CD}{\sin(24)} \\ \frac{15052.07}{\sin(90)} \cdot \sin(24)= \frac{CD}{\sin(24)}\cdot \sin(24) \\\frac{15052.07}{\sin(90)} \cdot \sin(24)= CD \]

  43. phi
    • 2 years ago
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    now round to the nearest foot. that is the answer for part (b)

  44. phi
    • 2 years ago
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    This was a difficult problem. But it go faster if you try to learn some of this.

  45. lala2
    • 2 years ago
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    Thank you @phi! I really apperciate your help and patience

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