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purplemexican

  • one year ago

Find all solutions for a triangle with A=30° a=4 b=8

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  1. mathstudent55
    • one year ago
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    |dw:1439394533674:dw|

  2. mathstudent55
    • one year ago
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    Law of sines: \(\large \dfrac{4}{\sin 30^o} = \dfrac{8}{\sin B} \) \(4 \sin B = 8 \sin 30^o\) \(\sin B = \dfrac{8 \sin 30^o}{4} \) \(\sin B = 1\) \(B = 90^o\) \(C = 180 - 30^o - 90^o = 60^o\) This means our drawing above is incorrect, ans we have a 30-60-90 triangle.

  3. mathstudent55
    • one year ago
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    |dw:1439395176899:dw|

  4. mathstudent55
    • one year ago
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    |dw:1439395554302:dw|

  5. mathstudent55
    • one year ago
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    Law of sines: \(\large \dfrac{4}{\sin 30^o} = \dfrac{c}{\sin 60^o} \) \(\large c = \dfrac{4 \sin 60^o}{\sin 30^o} \) \(\large c = 6.93\) Of course, we could have solved for c using the ratio of the lengths of the sides of a 30-60-90 triangle. \(1 : \sqrt 3 : 2\) \(4 : c : 2\) \(\dfrac{1}{\sqrt 3} = \dfrac{4}{c} \) \(c = 4 \sqrt 3 \approx 6.93\)

  6. Purplemexican
    • one year ago
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    thank you

  7. mathstudent55
    • one year ago
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    You're welcome.

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