Prove: Δ/s-a = s*tan a/2
\[\frac{ Δ }{ s-a } = s \tan \frac{ \alpha }{ 2 }\]
using R.H.S ??
\[\tan \frac{ \alpha }{ 2 } = \sqrt{\frac{ (s-b)(s-c) }{ s(s-a) }}\]
and \[\Delta = \sqrt{s(s-a)(s-b)(s-c)}\]
then its direct. \(\dfrac{\Delta}{s-a} =\dfrac{ \sqrt{s(s-a)(s-b)(s-c)}}{s-a}=\dfrac{ \sqrt{s(s-b)(s-c)}}{\sqrt{s-a}}\)
\(s\tan \frac{ \alpha }{ 2 } =s \sqrt{\frac{ (s-b)(s-c) }{ s(s-a) }}= \sqrt{\frac{s (s-b)(s-c) }{ (s-a) }}\)
LHS = RHS.
we have to prove it using a single side
okk.
i think we should choose R.H.S
any1 can be chosen., \(\dfrac{\Delta}{s-a} =\dfrac{ \sqrt{s(s-a)(s-b)(s-c)}}{s-a}\\ =\dfrac{ \sqrt{s(s-b)(s-c)}}{\sqrt{s-a}} \\=\sqrt{\frac{s (s-b)(s-c) }{ (s-a) }} \\ =s \sqrt{\frac{ (s-b)(s-c) }{ s(s-a) }} \\ = s\tan \frac{ \alpha }{ 2 } \)
WHAT HAPPENED IN THE SECOND STEP WITH s-a ???
\(\dfrac{\sqrt{s-a}}{s-a}=\dfrac{1}{\sqrt{s-a}}\)
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