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Question Number 21889 by ajfour last updated on 06/Oct/17

If a,b, and A of a triangle are fixed and two possible values of the third side be c_1 and c_2 such that c_1 ^2 +c_1 c_2 +c_2 ^2 =a^2 , then find angle A.

$${If}\:{a},{b},\:{and}\:{A}\:{of}\:{a}\:{triangle}\:\:{are}\: \\ $$$${fixed}\:{and}\:{two}\:{possible}\:{values}\:{of}\:{the}\: \\ $$$${third}\:{side}\:{be}\:{c}_{\mathrm{1}} {and}\:{c}_{\mathrm{2}} {such}\:{that} \\ $$$$\boldsymbol{{c}}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{{c}}_{\mathrm{1}} \boldsymbol{{c}}_{\mathrm{2}} +\boldsymbol{{c}}_{\mathrm{2}} ^{\mathrm{2}} =\boldsymbol{{a}}^{\mathrm{2}} ,\:{then}\:{find}\:{angle}\:{A}. \\ $$

Answered by mrW1 last updated on 06/Oct/17

a^2 =b^2 +c^2 −2bc×cos A ⇒c^2 −(2b cos A)c+(b^2 −a^2 )=0 c_1 +c_2 =2b cos A c_1 c_2 =b^2 −a^2 c_1 ^2 +c_1 c_2 +c_2 ^2 =a^2 c_1 ^2 +2c_1 c_2 +c_2 ^2 =a^2 +c_1 c_2 (c_1 +c_2 )^2 =a^2 +c_1 c_2 ⇒4b^2 cos^2 A=a^2 +b^2 −a^2 =b^2 ⇒4 cos^2 A=1 ⇒cos A=±(1/2) ⇒A=60° or 120°

$$\mathrm{a}^{\mathrm{2}} =\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{2bc}×\mathrm{cos}\:\mathrm{A} \\ $$$$\Rightarrow\mathrm{c}^{\mathrm{2}} −\left(\mathrm{2b}\:\mathrm{cos}\:\mathrm{A}\right)\mathrm{c}+\left(\mathrm{b}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\mathrm{c}_{\mathrm{1}} +\mathrm{c}_{\mathrm{2}} =\mathrm{2b}\:\mathrm{cos}\:\mathrm{A} \\ $$$$\mathrm{c}_{\mathrm{1}} \mathrm{c}_{\mathrm{2}} =\mathrm{b}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \\ $$$$ \\ $$$$\boldsymbol{{c}}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{{c}}_{\mathrm{1}} \boldsymbol{{c}}_{\mathrm{2}} +\boldsymbol{{c}}_{\mathrm{2}} ^{\mathrm{2}} =\boldsymbol{{a}}^{\mathrm{2}} \\ $$$$\boldsymbol{{c}}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2}\boldsymbol{{c}}_{\mathrm{1}} \boldsymbol{{c}}_{\mathrm{2}} +\boldsymbol{{c}}_{\mathrm{2}} ^{\mathrm{2}} =\boldsymbol{{a}}^{\mathrm{2}} +\mathrm{c}_{\mathrm{1}} \mathrm{c}_{\mathrm{2}} \\ $$$$\left(\mathrm{c}_{\mathrm{1}} +\mathrm{c}_{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} +\mathrm{c}_{\mathrm{1}} \mathrm{c}_{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4b}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\mathrm{A}=\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} =\mathrm{b}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{A}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{A}=\pm\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{A}=\mathrm{60}°\:\mathrm{or}\:\mathrm{120}° \\ $$

Commented by ajfour last updated on 06/Oct/17

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

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