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anonymous

  • 4 years ago

can someone explain how to find values of the six trigonometric functions when given: "Θ lies in Quadrant II. tan(Θ)=-24/7"

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  1. anonymous
    • 4 years ago
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    what the answer?

  2. anonymous
    • 4 years ago
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    yeah, i'm not sure how to find it and i have four other problems for homework like it so i was wondering if someone could explain. do you knw?

  3. anonymous
    • 4 years ago
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    u want the value of \[\Theta\]??

  4. anonymous
    • 4 years ago
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    no i need to find the other six trigonometric functions by using the function of tan(Θ)

  5. Mertsj
    • 4 years ago
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    |dw:1328543195617:dw|

  6. Mertsj
    • 4 years ago
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    \[\sin \theta=\frac{side opposite}{hypotenuse}=\frac{24}{25}\]

  7. Mertsj
    • 4 years ago
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    \[\cos \theta=\frac{side adjacent}{hypotenuse}=\frac{-7}{25}\]

  8. anonymous
    • 4 years ago
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    you just use that with all of the functions? i have them written down. (: but how did you get 25?

  9. Mertsj
    • 4 years ago
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    \[\tan \theta=\frac{-24}{7}\]

  10. Mertsj
    • 4 years ago
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    I got 25 by using the Pythagorean Theorem

  11. Mertsj
    • 4 years ago
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    now the other functions are the reciprocals of those already given:

  12. Mertsj
    • 4 years ago
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    \[\cot \theta=\frac{1}{\tan \theta}=\frac{-7}{24}\]

  13. Mertsj
    • 4 years ago
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    \[\sec \theta=\frac{1}{\cos \theta}=\frac{-25}{7}\]

  14. Mertsj
    • 4 years ago
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    \[\csc \theta=\frac{1}{\sin \theta}=\frac{24}{25}\]

  15. anonymous
    • 4 years ago
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    thank you ! i'm still a little confuzed about the 25. \[a ^{2}+b ^{2}=c ^{2}\] ? right?

  16. Mertsj
    • 4 years ago
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    yes

  17. anonymous
    • 4 years ago
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    OH i'm sorry i get it now. i wasn't square rooting the final answer. thank you so much !

  18. Mertsj
    • 4 years ago
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    \[7^2+24^2=c^2=49+576=625\]

  19. Mertsj
    • 4 years ago
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    \[c^2=625\]

  20. Mertsj
    • 4 years ago
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    \[c=\sqrt{625}=25\]

  21. anonymous
    • 4 years ago
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    thank you!

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