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anonymous

  • one year ago

Exact measures

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  1. anonymous
    • one year ago
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  2. campbell_st
    • one year ago
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    so do you know able the exact value ratios in trigonometry...?

  3. anonymous
    • one year ago
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    yes the sin, cos and tan

  4. anonymous
    • one year ago
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    I get the proportions mixed up!

  5. anonymous
    • one year ago
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    for a i got 5sin30

  6. Compassionate
    • one year ago
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    @campbell_st , SOHCAHTOA isn't really a ratio. SOH = Sine OVER Hypotenuse CAH = Cosine over Hypotenuse TOA = Tanget over Adjacent

  7. campbell_st
    • one year ago
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    ok... so that's the easiest method... use the scaling factor of similar triangles... well here is the basic triangle |dw:1443990164897:dw| apply the scaling factor to the other sides to find the exact measures.

  8. ybarrap
    • one year ago
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    You can use law of sines: $$ \cfrac{a}{\sin C}=\cfrac{b}{\sin B}\\ \cfrac{5}{\sin 30^\circ }=\cfrac{b}{\sin 60^\circ} \\ $$ Solve for b Make sense?

  9. campbell_st
    • one year ago
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    so remember \[2.5 = \frac{5}{2}\] so multiply by the improper fraction rather than the decimal and then you don't have to worry about trig ratios

  10. Mertsj
    • one year ago
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    |dw:1443990313447:dw|

  11. ybarrap
    • one year ago
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    Use $$ \sin 60^\circ =\cfrac{\sqrt 3}{2}\\ \sin 30^\circ =\cfrac{ 1}{2}\\ $$

  12. campbell_st
    • one year ago
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    this has become to confusing for me... I'll let others answer the question rather than bombard you with options

  13. ybarrap
    • one year ago
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    *correction $$ \cfrac{a}{\sin \color{red} A}=\cfrac{b}{\sin B}\\ \cfrac{5}{\sin 30^\circ }=\cfrac{b}{\sin 60^\circ} \\ $$

  14. anonymous
    • one year ago
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    ybarrap, how would I solve based off of the proportion.. it's a little confusing for me

  15. Mertsj
    • one year ago
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    I see from this question and the previous question you posted that you are studying the special right triangles. You are supposed to learn the relationships between the sides and angles of these special triangles so that you never have to calculate them again.

  16. Compassionate
    • one year ago
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    Lets say you have a triangle with Hypotenuse x, adjacent y, and opposite z. say I'm given the hypotenuse x's degree of 32 degrees, and I'm given an adjacent angle 10. Now all I can solve for is either the hypotenuse or opposite angle. So, if my test question said: find the hypotenuse I would say: Tan(32) = Adjacent/x Or: Tan(32) = 10/x

  17. Mertsj
    • one year ago
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    Here's the deal: In the 45-45-90 isosceles triangle, the legs are equal and the hypotenuse is the leg multiplied by sqrt(2) In the 30-60-90 triangle, the leg opposite the 30 degree angle is 1/2 the hypotenuse and the other leg is the short leg multiplied by sqrt(3)

  18. Mertsj
    • one year ago
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    You should memorize those two facts and practice enough that you can apply them automatically without using trig ratios or the law of sines or any of the rest of it.

  19. Compassionate
    • one year ago
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    @Mertsj , @campbell_st , @ybarrap -- I anonymously elect @Mertsj to help this student -- as we are confusing her with our differences in methods.

  20. campbell_st
    • one year ago
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    I bailed early because of the flood of information

  21. anonymous
    • one year ago
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    Mertsj, it's more easier for me when I actually set up proportions

  22. Mertsj
    • one year ago
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    Do not take the easy way out, sweetheart. Trust the voice of experience. If you memorize these two special triangles, it will be a great benefit to you in the future.

  23. anonymous
    • one year ago
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    ok :) now that I wrote out the facts, how would that help me? I'm just so used to writing out proportions

  24. Nnesha
    • one year ago
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    according to the previous question: you should include sin cos and tan ratios in your answer

  25. anonymous
    • one year ago
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    Nnesha, I'm not sure if i'm setting up the proportion up for a correctly. I wrote a=sin30

  26. anonymous
    • one year ago
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    a=5sin30*

  27. phi
    • one year ago
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    It might help to look at an intro to trig https://www.khanacademy.org/math/trigonometry/basic-trigonometry/basic_trig_ratios/v/basic-trigonometry that will help you figure out the ratios

  28. phi
    • one year ago
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    but yes sin 30 = a/c and c is 5 , so sin 30 = a/5 multiply both sides by 5 \[ 5\sin 30 =5\cdot \frac{a}{5}\\ 5\sin 30 =a \]

  29. phi
    • one year ago
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    people memorize sin,cos, tan for the angles 0, 30, 45, 60, and 90 that is, you (should) know sin 30 = 1/2 so a= 5* sin 30 = 5* 1/2 = 5/2

  30. phi
    • one year ago
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    That is using trig. For "special triangles" 30-60-90 and 45-45-90 people memorize some "short cuts"

  31. phi
    • one year ago
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    Are you studying trig or "special triangles" ?

  32. anonymous
    • one year ago
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    ok thank you :) for the length of b I got 5sqrt(3)/2 not sure if it's correct

  33. anonymous
    • one year ago
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    I'm studying trig

  34. phi
    • one year ago
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    yes, b is \(5\sqrt3/2\) cos 30 = b/c sin 60 = b/c either way, with c=5, you will get b= 5 sqr(3)/2

  35. anonymous
    • one year ago
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    got it thank you phi!

  36. phi
    • one year ago
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    if you have time, get some popcorn and watch Khan's videos

  37. anonymous
    • one year ago
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    thanks for the advice :D definitely will do!

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