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anonymous

  • one year ago

In triangle ABC , m angle A=41 degrees, m angle B=32 degrees and AC= 9 inches what is AB to the nearest tenth of an inch a. 13.1 in b. 13.6 in c. 16.2 in d. 16.9 in

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  1. anonymous
    • one year ago
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    |dw:1432846650149:dw| I'm assuming it looks like this?

  2. anonymous
    • one year ago
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    i guess? it didnt have a picture on this one

  3. johnweldon1993
    • one year ago
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    So this can just be solved with the law of sines You have your triangle *drawn above* and you have the length of 1 other side So we can use \[\large \frac{sin(32)}{9} = \frac{sin(107)}{AB}\] And solve for AB

  4. anonymous
    • one year ago
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    ok and i solve for AB how? @johnweldon1993

  5. johnweldon1993
    • one year ago
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    Wait, first, have you seen this before? The law of sines?

  6. anonymous
    • one year ago
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    can you walk me through this? or would that be asking for to much? @johnweldon1993 and no i changed schools and skiped like 3 units so im trying to get all of this

  7. johnweldon1993
    • one year ago
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    No never asking too much :) but first, what class is the for? I just dont want to introduce something you wouldnt even be learning :)

  8. anonymous
    • one year ago
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    Algebra 2

  9. johnweldon1993
    • one year ago
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    Okay then yeah this applies :) Okay so...in general You have a triangle |dw:1432852105182:dw|

  10. johnweldon1993
    • one year ago
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    So first we need to solve for angle C We know that the interior angles of a triangle always add to 180 So since we have 41...and 32...we have 73 of those 180 degrees So angle C must be the remaining 107 degrees right?

  11. anonymous
    • one year ago
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    ok...

  12. johnweldon1993
    • one year ago
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    does that make sense? Gotta make sure you're staying with me :)

  13. anonymous
    • one year ago
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    yes im with you

  14. johnweldon1993
    • one year ago
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    So now, we apply the law of sines What THAT states...is: \[\large \frac{sin(A)}{a} = \frac{sin(B)}{b} = \frac{sin(C)}{c}\] Where A,B and C are the angle measurements we have and a,b and c are the side lengths we have

  15. johnweldon1993
    • one year ago
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    Now the way we use it....is if we have angle A, then side "A" would be the side the angle faces |dw:1432852557818:dw|

  16. anonymous
    • one year ago
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    that's familiar

  17. johnweldon1993
    • one year ago
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    So...since you have side B right...here it would be AC |dw:1432852698148:dw| And you want to solve for side C which here would be AB |dw:1432852734405:dw| we can write \[\large \frac{sin(\text{Angle of B)}}{\text{Length of side B = 9}} = \frac{sin(\text{angle of C)}}{\text{length of side C = ???}}\] \[\large \frac{sin(32)}{9} = \frac{sin(107)}{AB}\]

  18. johnweldon1993
    • one year ago
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    And then we just solve for AB Now to do that...we just cross multiply \[\large ABsin(32) = 9sin(107)\] And divide both sides by the sin(32) so \[\large AB = \frac{9sin(107)}{sin(32)}\]

  19. anonymous
    • one year ago
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    ok.. um sorry its taking a while for me to catch on

  20. johnweldon1993
    • one year ago
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    No that's fine :) just tell me if you want me to explain a part over :)

  21. anonymous
    • one year ago
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    it was pretty explanatory just takes a lot of practice thank you so much!

  22. johnweldon1993
    • one year ago
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    Of course :)

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