-Welp-
  • -Welp-
How do I solve this? http://i.imgur.com/UOhG3B8.png (trig help)
Geometry
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-Welp-
  • -Welp-
How do I solve this? http://i.imgur.com/UOhG3B8.png (trig help)
Geometry
schrodinger
  • schrodinger
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Study_together
  • Study_together
@TheSmartOne can help you. I have no clue how to do this
Astrophysics
  • Astrophysics
What are we solving for x, or y or both, use your trig ratios.
Astrophysics
  • Astrophysics
|dw:1436505609850:dw|

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Astrophysics
  • Astrophysics
What ratio can we use to find the x, |dw:1436505727553:dw|
-Welp-
  • -Welp-
I'm trying to solve for both x and y. "What ratio can we use to find the x, " Tan X=\[6\sqrt{3} ?\]
Astrophysics
  • Astrophysics
Yeah tan is good! So we have \[\tan \theta = \frac{ opp }{ adj } = \frac{ 18 }{ x }\] \[x = \frac{ 18 }{ \tan \theta } = \frac{ 18 }{ \tan(31) }\]
Astrophysics
  • Astrophysics
After you have x, you can find y using Pythagorean or using the ratios :)
-Welp-
  • -Welp-
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.
Astrophysics
  • Astrophysics
You shouldn't have square root?
Astrophysics
  • Astrophysics
|dw:1436519818790:dw| that's the square root symbol, what are you trying to do?
-Welp-
  • -Welp-
Nevermind. Y=\[3\sqrt{38}\]?
Astrophysics
  • Astrophysics
\[\sin \theta = \frac{ opp }{ hyp } = \frac{ 18 }{ y } \implies y = \frac{ 18 }{ \sin(31) } \]
Astrophysics
  • Astrophysics
You should also consider your answers, hypotenuse is the longest side, would your answer really make sense?
-Welp-
  • -Welp-
\[18^{2} + 6\sqrt{3}^{2} 342 \sqrt{342}= 3\sqrt{38}\] ^I was trying to solve it with the Pythagorean theorem. I don't know where I screwed up 18/sin(31)=34.9
Astrophysics
  • Astrophysics
\[a^2+b^2 = c^2 \implies 18^2 + (29.957...)^2 = y^2 \implies y = \sqrt{18^2+(29.957...)^2} =34.948...\]

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