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AonZ

  • 3 years ago

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

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  1. AonZ
    • 3 years ago
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    |dw:1367018505450:dw|

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

  3. AonZ
    • 3 years ago
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    @hartnn help please!!!

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

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

  6. Mertsj
    • 3 years ago
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    |dw:1367022862806:dw|

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

  8. Mertsj
    • 3 years ago
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    Just plug into the formula.

  9. AonZ
    • 3 years ago
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    yes!!! that actually worked!! Thank you so much

  10. Mertsj
    • 3 years ago
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    yw

  11. hartnn
    • 3 years ago
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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

  12. AonZ
    • 3 years ago
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    ohhhh i see

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