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AonZ

HELP PLEASE!!! ABCD is a cyclic quadrilateral in which AB=5. BC=6. CD=7. AD=8. show that cos ADC = 13/43

  • 11 months ago
  • 11 months ago

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  1. AonZ
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    |dw:1367018505450:dw|

    • 11 months ago
  2. AonZ
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    @kropot72 @terenzreignz @Mertsj Help pls Btw i know angles D+B = 180

    • 11 months ago
  3. AonZ
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    @hartnn help please!!!

    • 11 months ago
  4. Mertsj
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    There is a formula for this. First let me relabel your picture so you will understand the formula better.

    • 11 months ago
  5. AonZ
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    ok... only formula i know is angle B + D=180 and angle A+C = 180

    • 11 months ago
  6. Mertsj
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    |dw:1367022862806:dw|

    • 11 months ago
  7. Mertsj
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    \[\cos A=\frac{a^2+d^2-b^2-c^2}{2(ad+bc)}\]

    • 11 months ago
  8. Mertsj
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    Just plug into the formula.

    • 11 months ago
  9. AonZ
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    yes!!! that actually worked!! Thank you so much

    • 11 months ago
  10. Mertsj
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    yw

    • 11 months ago
  11. hartnn
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    did you get how how found cos A ? he just applied cosine rule, twice. once in triangle ABD and then in triangle BCD, also knowing A+C = 180, so cos C = -cos A

    • 11 months ago
  12. AonZ
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    ohhhh i see

    • 11 months ago
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