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anonymous

  • one year ago

I really need help with this please. :) Here is a picture of the problem: https://cdn.ple.platoweb.com/EdAssets/2aa74bea51264416af45cb620a392471?ts=635545843843870000 If a = 24 mm, c = 26 mm, and m∠B = 33°, what is the approximate area of ABC? 1. 261.67 mm2 2. 54.46 mm2 3. 169.93 mm2 4. 339.85 mm2

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  1. SolomonZelman
    • one year ago
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    This is a right triangle?

  2. anonymous
    • one year ago
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    I think so.

  3. SolomonZelman
    • one year ago
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    Ok, then you can use the Pythagorean theorem to find the side b.

  4. anonymous
    • one year ago
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    I would do that, but it is a degree instead of a number.

  5. SolomonZelman
    • one year ago
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    I am saying to use the `. a² + b² = c² . `

  6. SolomonZelman
    • one year ago
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    24²+b²=26²

  7. anonymous
    • one year ago
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    Okay. Thank you so much!

  8. SolomonZelman
    • one year ago
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    we aren'tdone yet, are we?

  9. SolomonZelman
    • one year ago
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    Did you find the side b?

  10. SolomonZelman
    • one year ago
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    (It will be a precise, natural number, answer)

  11. anonymous
    • one year ago
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    I'm working on it.

  12. SolomonZelman
    • one year ago
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    Ok, sure, take your time:)

  13. anonymous
    • one year ago
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    I got 10.

  14. SolomonZelman
    • one year ago
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    Oh, very good. that is right!

  15. SolomonZelman
    • one year ago
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    |dw:1441584583951:dw|

  16. SolomonZelman
    • one year ago
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    So you can already tell which choice is closer to the area.

  17. anonymous
    • one year ago
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    24 right?

  18. jdoe0001
    • one year ago
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    |dw:1441585425017:dw|

  19. SolomonZelman
    • one year ago
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    I wonder that they do not have the answer choice for that....

  20. anonymous
    • one year ago
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    So, the test makers didn't put the correct choice on it?

  21. SolomonZelman
    • one year ago
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    No, I think it is rather our mistake that we assumed that your triangle is a right triangle.

  22. jdoe0001
    • one year ago
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    hmmm

  23. anonymous
    • one year ago
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    Okay, so how would we solve this now?

  24. SolomonZelman
    • one year ago
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    I think we need to apply the law of cosines. \(b^2=a^2+c^2-2ac{\rm Cos}(B)\)

  25. jdoe0001
    • one year ago
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    have you covered the law of cosines yet?

  26. anonymous
    • one year ago
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    Not yet, but this is supposed to be a Algebra 1 study.

  27. SolomonZelman
    • one year ago
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    https://www.mathsisfun.com/algebra/trig-cosine-law.html

  28. jdoe0001
    • one year ago
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    hmmm so.. haemm I think is safe to assume then, that is a right triangle

  29. SolomonZelman
    • one year ago
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    This is a good link. mathisfun is always easy to read:)

  30. SolomonZelman
    • one year ago
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    yes jdoe, I assumed that the first I encountered it, but there is no option of 120.

  31. jdoe0001
    • one year ago
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    well, right... I'd use the law of cosines, but if your teacher hasn't covered, there's no sense in giving you this exericse yet

  32. anonymous
    • one year ago
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    Thank you for the link, I will check it out.

  33. jdoe0001
    • one year ago
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    law of cosines and Heron's formula to get the area then again, you may not have covered heron's formula either you might think the formula has a two long legs, but it doesn't =)

  34. SolomonZelman
    • one year ago
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    \(b^2=c^2+a^2-2(a)(c){\rm Cos}(B)\) \(b^2=26^2+24^2-2(24)(26){\rm Cos}(33)\) \(b=\sqrt{26^2+24^2-2(24)(26){\rm Cos}(33)}=14.33\)

  35. SolomonZelman
    • one year ago
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    |dw:1441585266641:dw|

  36. anonymous
    • one year ago
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    I got 171.96 as the area, but that is not one of the answers either.

  37. SolomonZelman
    • one year ago
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    I am trying to find every thing in the triangle, perhaps, before doing area.

  38. SolomonZelman
    • one year ago
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    \(\sin(33)/14.33{~~}={~~}\sin(C)/26\)

  39. SolomonZelman
    • one year ago
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    |dw:1441585636236:dw|

  40. SolomonZelman
    • one year ago
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    So the height is less that 14, but not much less than 14.33. We can guess it is 12, then it is approaches choice A.

  41. SolomonZelman
    • one year ago
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    (That is if you want to just approximate the area. But we can try to find the height.... which is a lot of sweating)

  42. anonymous
    • one year ago
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    Okay, so is this Algebra, it seems more like Geometry to me?

  43. anonymous
    • one year ago
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    Thank you guys so much. Have a great day. :)

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