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

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

$$\:{let}\:{ABC}\:\:{be}\:{a}\:{triangle}\:{AB}={c}\:\:{AC}={b}\:\:{BC}={a} \\ $$$${Show}\:{that}\:{ABC}\:{is}\:{right}\:\Leftrightarrow\:\:{tan}\left(\frac{{B}}{\mathrm{2}}\right)=\frac{{a}+{c}}{{b}}\: \\ $$

Commented by 1549442205PVT last updated on 14/Oct/20

Question is′nt exact.Example,ΔABC A=90° ,B=60°,a=BC=2⇒AB=c=2.sin30=1 AC=b=2.sin60°=(√3) tan(B/2)=tan30°=(1/( (√3)))≠((a+c)/b)=((2+1)/( (√3)))=(√3) You should look at your question

$$\mathrm{Question}\:\mathrm{is}'\mathrm{nt}\:\mathrm{exact}.\mathrm{Example},\Delta\mathrm{ABC} \\ $$$$\mathrm{A}=\mathrm{90}° \\ $$$$\:,\mathrm{B}=\mathrm{60}°,\mathrm{a}=\mathrm{BC}=\mathrm{2}\Rightarrow\mathrm{AB}=\mathrm{c}=\mathrm{2}.\mathrm{sin30}=\mathrm{1} \\ $$$$\mathrm{AC}=\mathrm{b}=\mathrm{2}.\mathrm{sin60}°=\sqrt{\mathrm{3}} \\ $$$$\mathrm{tan}\frac{\mathrm{B}}{\mathrm{2}}=\mathrm{tan30}°=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\neq\frac{\mathrm{a}+\mathrm{c}}{\mathrm{b}}=\frac{\mathrm{2}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}=\sqrt{\mathrm{3}} \\ $$$$\mathrm{You}\:\mathrm{should}\:\mathrm{look}\:\mathrm{at}\:\mathrm{your}\:\mathrm{question} \\ $$

Commented by snipers237 last updated on 14/Oct/20

Forgive sir , it was cotan((B/2))=((a+c)/b)

$${Forgive}\:{sir}\:,\:{it}\:{was}\:\:{cotan}\left(\frac{{B}}{\mathrm{2}}\right)=\frac{{a}+{c}}{{b}}\: \\ $$

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