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koli123able

  • 3 years ago

Prove the followings

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  1. koli123able
    • 3 years ago
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    Number 1 \[\frac{ \Delta }{ s-a }=s \tan \frac{ \alpha }{ 2 }\]

  2. mathslover
    • 3 years ago
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    what does alpha , a , s , \(\Delta\) represent ?

  3. koli123able
    • 3 years ago
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    \[\Delta=\sqrt{s(s-a)(s-b)(s-c)}\]

  4. koli123able
    • 3 years ago
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    \[\tan \frac{ \alpha }{ 2 }=\sqrt{\frac{ (s-b)(s-c) }{ s(s-a) }}\]

  5. mathslover
    • 3 years ago
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    right... got it now..

  6. mathslover
    • 3 years ago
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    so now put these values : \[\large{\frac{ \sqrt{s(s-a)(s-b)(s-c)}}{s-a} = \frac{\sqrt{s-a} \sqrt{s(s-b)(s-c)}}{s-a} }\]

  7. koli123able
    • 3 years ago
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    prove it using either l.h.s or r.h.s

  8. mathslover
    • 3 years ago
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    Yeah I am using LHS only

  9. koli123able
    • 3 years ago
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    okay.. :)

  10. mathslover
    • 3 years ago
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    well I can write :; \(\large{\frac{\sqrt{s-a}}{s-a} }\) as \(\frac{1}{\large{\sqrt{s-a}}}\)

  11. mathslover
    • 3 years ago
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    therefore I get : \[\large{\frac{\sqrt{s(s-b)(s-c)}}{\sqrt{s-a}}}\] = \[\large{\sqrt{\frac{(s-b)(s-c)s^2}{s(s-a)}}} \] That is : \[\large{s\tan \frac{\alpha}{2}}\]

  12. mathslover
    • 3 years ago
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    got it ?

  13. koli123able
    • 3 years ago
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    can u prove it by multiplying and dividing trick i did'nt understand the roots u changed

  14. koli123able
    • 3 years ago
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    like taking r.h.s and M & D it by \[\sqrt{s(s-a)}\]

  15. mathslover
    • 3 years ago
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    see : |dw:1360917633731:dw|

  16. koli123able
    • 3 years ago
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    NUMBER 2\[\frac{ s-a }{ \Delta }+\frac{ s-b }{ \Delta }+\frac{ s-c }{ \Delta }= \frac{ s }{ \Delta }\]

  17. koli123able
    • 3 years ago
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    Using lhs

  18. koli123able
    • 3 years ago
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    \[\frac{ s-a+s-b+s-c }{ \Delta }\]

  19. koli123able
    • 3 years ago
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    what to do after this??

  20. hartnn
    • 3 years ago
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    didn't you ask the first one earlier also? :O also, for 2nd, we know that s is the semi-perimeter , hence, \(s= (a+b+c)/2 ------> 2s = a+b+c\) so, your numerator\( = s-a+s-b+s-c = 3s- (a+b+c) =.... ?\)

  21. koli123able
    • 3 years ago
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    :D yes i did asked the first 1

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