let ABC be a triangle AB=c AC=b BC=a Show that ABC is right ⇔ tan((B/2))=((a+c)/b)


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Question Number 117818 by snipers237 last updated on 13/Oct/20

$l e t A B C b e a t r i a n g l e A B = c A C = b B C = a$ $S h o w t h a t A B C i s r i g h t \Leftrightarrow t a n (\frac{B}{2}) = \frac{a + c}{b}$
Commented by 1549442205PVT last updated on 14/Oct/20

$Question {is}^{'} nt exact . Example, Δ ABC$ $A = 90 °$ $, B = 60 °, a = BC = 2 \Rightarrow AB = c = 2 . \sin 30 = 1$ $AC = b = 2 . \sin 60 ° = \sqrt{3}$ $\tan \frac{B}{2} = \tan 30 ° = \frac{1}{\sqrt{3}} \neq \frac{a + c}{b} = \frac{2 + 1}{\sqrt{3}} = \sqrt{3}$ $You should look at your question$
Commented by snipers237 last updated on 14/Oct/20

$F o r g i v e s i r, i t w a s c o t a n (\frac{B}{2}) = \frac{a + c}{b}$
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