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anonymous

  • one year ago

Solve the triangle. A = 50°, b = 13, c = 6

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  1. DanJS
    • one year ago
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    |dw:1438968599375:dw|

  2. anonymous
    • one year ago
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    Yes I get that but how do I find my missing pieces

  3. DanJS
    • one year ago
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    You can use the law of sines or cosines

  4. anonymous
    • one year ago
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    I don't get how to do that I'm going virtual school and I need someone to walk me step by step

  5. DanJS
    • one year ago
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    http://www.mathwarehouse.com/trigonometry/law-of-cosines-formula-examples.php

  6. DanJS
    • one year ago
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    \[a^2 = 6^2 + 13^2 - 2(6)(13)*\cos(50)\]

  7. anonymous
    • one year ago
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    I don't have a calculator except my phone so when I do it, I don't get any of my answer choices

  8. DanJS
    • one year ago
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    you get side a from that one, then you can use law of sines to get the last 2 angles \[\frac{ \sin(A) }{ a } = \frac{ \sin(B) }{ b } = \frac{ \sin(C) }{ c }\]

  9. anonymous
    • one year ago
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    Is there anyway you can calculate that because my phone doesn't get me the correct answer

  10. DanJS
    • one year ago
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    a^2 = 36 + 169 - 156*(0.6428) a^2 = 104.7 a = about 10.2

  11. DanJS
    • one year ago
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    \[\frac{ \sin(50) }{ 10.2 }=\frac{ \sin(B) }{ 13 }\]

  12. DanJS
    • one year ago
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    B= about 77.5 degrees

  13. DanJS
    • one year ago
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    THe last angle is 180 - the other 2

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