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Derive tan(A/2) = sqrt[(s-b)(s-c)/{s(s-a)}]
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Enuyvatar
Nov 23, 2009, 12:15:56 PM11/23/09
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How does one derive
tan(A/2) = sqrt[(s-b)(s-c)/{s(s-a)}]?
tan(A/2) = sqrt[(s-b)(s-c)/{s(s-a)}]?
Thanks.
Enuyvatar
Nov 23, 2009, 1:20:25 PM11/23/09
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Solved.
basuki
Dec 18, 2009, 8:17:16 AM12/18/09
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> How does one derive
> tan(A/2) = sqrt[(s-b)(s-c)/{s(s-a)}]?
>
> Thanks.
What did you mean with A,a,b,c, and s?> tan(A/2) = sqrt[(s-b)(s-c)/{s(s-a)}]?
>
> Thanks.
It is usual to us if there is a triangle ABC,
a is the side in front of angle A,
b is the side in front of angle B,
c is the side in front of angle C
And s=(a+b+c)/2.
Was what you meant as above?
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