Menu Close

let-ABC-be-a-triangle-AB-c-AC-b-BC-a-Show-that-ABC-is-right-tan-B-2-a-c-b-




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

Leave a Reply

Your email address will not be published. Required fields are marked *