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JoãoVitorMC

  • 3 years ago

Consider the isosceles triangle with AB = AC. Suppose the angle bisector of B intersects the side AC at D and that BC = BD + AD. Determine the angle A.

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  1. JoãoVitorMC
    • 3 years ago
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    |dw:1350605708301:dw|

  2. CliffSedge
    • 3 years ago
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    Wanna use trig, or regular plane geometry?

  3. JoãoVitorMC
    • 3 years ago
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    the easiest way to solve this

  4. CliffSedge
    • 3 years ago
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    I was looking at this last night. I'm trying to find my notes on it.

  5. JoãoVitorMC
    • 3 years ago
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    ok, i'm trying to solve this, but this one is really hard...

  6. CliffSedge
    • 3 years ago
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    Yeah, I'm trying combinations of law of cosines and law of sines.

  7. JoãoVitorMC
    • 3 years ago
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    but nothing yet

  8. estudier
    • 3 years ago
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    |dw:1350693786588:dw|

  9. estudier
    • 3 years ago
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    I will have to make a better drawing.....Answer s/b 100, I think....

  10. estudier
    • 3 years ago
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    |dw:1350694614179:dw|

  11. JoãoVitorMC
    • 3 years ago
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    what is "s/b 100"?

  12. estudier
    • 3 years ago
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    should be

  13. JoãoVitorMC
    • 3 years ago
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    ohh ok i got it

  14. JoãoVitorMC
    • 3 years ago
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    thank you so much!

  15. estudier
    • 3 years ago
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    yw:-)

  16. JoãoVitorMC
    • 3 years ago
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    hey where you get this 180° -5(Betha) = 4(Betha) ?

  17. estudier
    • 3 years ago
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    ADB is 3 Beta

  18. JoãoVitorMC
    • 3 years ago
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    yes but i still don't see how to get 4 Beta...

  19. estudier
    • 3 years ago
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    look at triangle AED

  20. estudier
    • 3 years ago
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    For 180-5 Beta.

  21. JoãoVitorMC
    • 3 years ago
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    yes that i got

  22. JoãoVitorMC
    • 3 years ago
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    but them you set this equal to 4 beta, then i'm lost...

  23. estudier
    • 3 years ago
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    BD = BE so 4 Beta

  24. JoãoVitorMC
    • 3 years ago
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    ohhhh now i see

  25. JoãoVitorMC
    • 3 years ago
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    thank you soo much!

  26. estudier
    • 3 years ago
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    yw:-)

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