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-Welp-

  • one year ago

How do I solve this? http://i.imgur.com/UOhG3B8.png (trig help)

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  1. Study_together
    • one year ago
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    @TheSmartOne can help you. I have no clue how to do this

  2. Astrophysics
    • one year ago
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    What are we solving for x, or y or both, use your trig ratios.

  3. Astrophysics
    • one year ago
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    |dw:1436505609850:dw|

  4. Astrophysics
    • one year ago
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    What ratio can we use to find the x, |dw:1436505727553:dw|

  5. -Welp-
    • one year ago
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    I'm trying to solve for both x and y. "What ratio can we use to find the x, " Tan X=\[6\sqrt{3} ?\]

  6. Astrophysics
    • one year ago
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    Yeah tan is good! So we have \[\tan \theta = \frac{ opp }{ adj } = \frac{ 18 }{ x }\] \[x = \frac{ 18 }{ \tan \theta } = \frac{ 18 }{ \tan(31) }\]

  7. Astrophysics
    • one year ago
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    After you have x, you can find y using Pythagorean or using the ratios :)

  8. -Welp-
    • one year ago
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    I don't know how to do those with square roots. I have no idea how to stop the square root symbol on my calculator.

  9. Astrophysics
    • one year ago
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    You shouldn't have square root?

  10. Astrophysics
    • one year ago
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    |dw:1436519818790:dw| that's the square root symbol, what are you trying to do?

  11. -Welp-
    • one year ago
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    Nevermind. Y=\[3\sqrt{38}\]?

  12. Astrophysics
    • one year ago
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    \[\sin \theta = \frac{ opp }{ hyp } = \frac{ 18 }{ y } \implies y = \frac{ 18 }{ \sin(31) } \]

  13. Astrophysics
    • one year ago
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    You should also consider your answers, hypotenuse is the longest side, would your answer really make sense?

  14. -Welp-
    • one year ago
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    \[18^{2} + 6\sqrt{3}^{2} 342 \sqrt{342}= 3\sqrt{38}\] ^I was trying to solve it with the Pythagorean theorem. I don't know where I screwed up 18/sin(31)=34.9

  15. Astrophysics
    • one year ago
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    \[a^2+b^2 = c^2 \implies 18^2 + (29.957...)^2 = y^2 \implies y = \sqrt{18^2+(29.957...)^2} =34.948...\]

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spraguer (Moderator)
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