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Question Number 15761 by Tinkutara last updated on 13/Jun/17

If in a ΔABC, ((2 cos A)/a) + ((cos B)/b) + ((2 cos C)/c) = (a/(bc)) + (b/(ac)) , prove that ∠A = 90°.

$$\mathrm{If}\:\mathrm{in}\:\mathrm{a}\:\Delta{ABC},\:\frac{\mathrm{2}\:\mathrm{cos}\:{A}}{{a}}\:+\:\frac{\mathrm{cos}\:{B}}{{b}}\:+\:\frac{\mathrm{2}\:\mathrm{cos}\:{C}}{{c}} \\ $$$$=\:\frac{{a}}{{bc}}\:+\:\frac{{b}}{{ac}}\:,\:\mathrm{prove}\:\mathrm{that}\:\angle{A}\:=\:\mathrm{90}°. \\ $$

Answered by ajfour last updated on 13/Jun/17

((2cos A)/a)+((cos B)/b)+((2cos C)/c)=((a^2 +b^2 )/(abc)) ⇒ 2bccos A+accos B+2abcos C =a^2 +b^2 knowing cos A=((b^2 +c^2 −a^2 )/(2bc)) cos B=((c^2 +a^2 −b^2 )/(2ac)) , cos C=((a^2 +b^2 −c^2 )/(2ab)) we get (b^2 +c^2 −a^2 )+(1/2)(c^2 +a^2 −b^2 )+ (a^2 +b^2 −c^2 ) =a^2 +b^2 ⇒ b^2 −a^2 +(1/2)(c^2 +a^2 −b^2 )=0 c^2 +a^2 −b^2 =2a^2 −2b^2 ⇒ b^2 +c^2 =a^2 hence a is the hypotenuse and ∠A = 90° .

$$\frac{\mathrm{2cos}\:{A}}{{a}}+\frac{\mathrm{cos}\:{B}}{{b}}+\frac{\mathrm{2cos}\:{C}}{{c}}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{abc}} \\ $$$$\Rightarrow\:\:\mathrm{2}{bc}\mathrm{cos}\:{A}+{ac}\mathrm{cos}\:{B}+\mathrm{2}{ab}\mathrm{cos}\:{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$${knowing}\:\mathrm{cos}\:{A}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}} \\ $$$$\:\:\mathrm{cos}\:{B}=\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ac}}\:\:,\:\:\mathrm{cos}\:{C}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$$${we}\:{get} \\ $$$$\:\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)\:\:\:\:\:={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{b}^{\mathrm{2}} −{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} \\ $$$$\:\Rightarrow\:\:\:\:{b}^{\mathrm{2}} +{c}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\:{hence}\:\boldsymbol{{a}}\:{is}\:{the}\:{hypotenuse}\:{and}\:\: \\ $$$$\:\:\:\angle{A}\:=\:\mathrm{90}°\:.\: \\ $$

Commented by Tinkutara last updated on 14/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Jun/17

((2sosA)/a)+((cosB)/b)+((2cosC)/c)= ((2bcosA+acosB)/(ab))+((2cosC)/c)= ((c+bcosA)/(ab))+((2cosC)/c)=((c^2 +bccosA+2abcosC)/(abc))= =((c^2 +b(c.cosA+acosC)+abcosC)/(abc))= =((c^2 +b^2 +abcosC)/(abc))=((2(b^2 +c^2 )+(a^2 +b^2 −c^2 ))/(2abc))= =((3b^2 +a^2 +c^2 )/(2abc))=(a/(bc))+(b/(ac))=((a^2 +b^2 )/(abc))⇒ 3b^2 +a^2 +c^2 =2a^2 +2b^2 ⇒b^2 +c^2 =a^2 .

$$\frac{\mathrm{2}{sosA}}{{a}}+\frac{{cosB}}{{b}}+\frac{\mathrm{2}{cosC}}{{c}}= \\ $$$$\frac{\mathrm{2}{bcosA}+{acosB}}{{ab}}+\frac{\mathrm{2}{cosC}}{{c}}= \\ $$$$\frac{{c}+{bcosA}}{{ab}}+\frac{\mathrm{2}{cosC}}{{c}}=\frac{{c}^{\mathrm{2}} +{bccosA}+\mathrm{2}{abcosC}}{{abc}}= \\ $$$$=\frac{{c}^{\mathrm{2}} +{b}\left({c}.{cosA}+{acosC}\right)+{abcosC}}{{abc}}= \\ $$$$=\frac{{c}^{\mathrm{2}} +{b}^{\mathrm{2}} +{abcosC}}{{abc}}=\frac{\mathrm{2}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)}{\mathrm{2}{abc}}= \\ $$$$=\frac{\mathrm{3}{b}^{\mathrm{2}} +{a}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}{abc}}=\frac{{a}}{{bc}}+\frac{{b}}{{ac}}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{abc}}\Rightarrow \\ $$$$\mathrm{3}{b}^{\mathrm{2}} +{a}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} \Rightarrow{b}^{\mathrm{2}} +{c}^{\mathrm{2}} ={a}^{\mathrm{2}} \:. \\ $$

Commented by Tinkutara last updated on 14/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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