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Question Number 209932 by cherokeesay last updated on 26/Jul/24
Answered by mahdipoor last updated on 26/Jul/24
BD^2 =100+49−140cosC=121+64−176cosA ((BD)/(sinC))=((BD)/(sinA))=2R ⇒C+A=180 ⇒149−140cosC=185−176cosA ⇒cosC=((185−149)/(−176−140))=−(9/(79)) ⇒2R=((√(149+140×(9/(79))))/( (√(1−((9/(79)))^2 ))))=12.93
$${BD}^{\mathrm{2}} =\mathrm{100}+\mathrm{49}−\mathrm{140}{cosC}=\mathrm{121}+\mathrm{64}−\mathrm{176}{cosA} \\ $$$$\frac{{BD}}{{sinC}}=\frac{{BD}}{{sinA}}=\mathrm{2}{R}\:\Rightarrow{C}+{A}=\mathrm{180} \\ $$$$\Rightarrow\mathrm{149}−\mathrm{140}{cosC}=\mathrm{185}−\mathrm{176}{cosA} \\ $$$$\Rightarrow{cosC}=\frac{\mathrm{185}−\mathrm{149}}{−\mathrm{176}−\mathrm{140}}=−\frac{\mathrm{9}}{\mathrm{79}} \\ $$$$\Rightarrow\mathrm{2}{R}=\frac{\sqrt{\mathrm{149}+\mathrm{140}×\frac{\mathrm{9}}{\mathrm{79}}}}{\:\sqrt{\mathrm{1}−\left(\frac{\mathrm{9}}{\mathrm{79}}\right)^{\mathrm{2}} }}=\mathrm{12}.\mathrm{93} \\ $$
Answered by a.lgnaoui last updated on 26/Jul/24
AB.AD=2R×h h=ABsin A1 R=((AD)/(2sin A1))=((8R)/( (√(4R^2 −121)))) 1=(8/( (√(4R^2 −121)))) 4R^2 =185 R =((√(185))/2) determinant (((AB.AD=2R.h h=ABsin A1)),( determinant ((( A1=∡ABD] )),(( R=6,8))))) Developement du resultat: BD^2 =185−176cos A A=(A1+A2)O { ((cos A1=((11)/(2R)) sin A1=(1/(2R))(√(4R^2 −121)))),((cos A2=(5/R) sin A2=(1/R)(√(R^2 −25)))) :} =149−140cos C ⇒ 36=176cos A+140cos A ⇒ 36=316cos A determinant ((),( determinant (((cos A )),(( (9/(79)) ))))) [1] sin A=sinA1cos A2+sin A2cos A1 cos A=cosA1 .cosA2 −sinA1 .sinA2 { () :}
$$\mathrm{AB}.\mathrm{AD}=\mathrm{2}\boldsymbol{\mathrm{R}}×\boldsymbol{\mathrm{h}} \\ $$$$\:\boldsymbol{\mathrm{h}}=\boldsymbol{\mathrm{AB}}\mathrm{sin}\:\boldsymbol{\mathrm{A}}\mathrm{1} \\ $$$$\:\:\:\boldsymbol{\mathrm{R}}=\frac{\boldsymbol{\mathrm{AD}}}{\mathrm{2sin}\:\boldsymbol{\mathrm{A}}\mathrm{1}}=\frac{\mathrm{8}\boldsymbol{\mathrm{R}}}{\:\sqrt{\mathrm{4}\boldsymbol{\mathrm{R}}^{\mathrm{2}} −\mathrm{121}}}\: \\ $$$$\mathrm{1}=\frac{\mathrm{8}}{\:\sqrt{\mathrm{4}\boldsymbol{\mathrm{R}}^{\mathrm{2}} −\mathrm{121}}}\:\: \\ $$$$ \\ $$$$ \\ $$$$\mathrm{4}\boldsymbol{\mathrm{R}}^{\mathrm{2}} =\mathrm{185}\:\:\:\:\:\boldsymbol{\mathrm{R}}\:\:\:\:\:=\frac{\sqrt{\mathrm{185}}}{\mathrm{2}} \\ $$$$\:\:\begin{array}{|c|c|}{\boldsymbol{\mathrm{AB}}.\boldsymbol{\mathrm{AD}}=\mathrm{2}\boldsymbol{\mathrm{R}}.\boldsymbol{\mathrm{h}}\:\:\:\boldsymbol{\mathrm{h}}=\boldsymbol{\mathrm{AB}}\mathrm{sin}\:\boldsymbol{\mathrm{A}}\mathrm{1}}\\{\begin{array}{|c|c|}{\left.\:\:\:\:\:\:\:\:\:\:\:\mathrm{A1}=\measuredangle\mathrm{ABD}\right]\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{R}}=\mathrm{6},\mathrm{8}}\\\hline\end{array}}\\\hline\end{array} \\ $$$$ \\ $$$$\boldsymbol{{Developement}}\:\boldsymbol{{du}}\:\boldsymbol{{resultat}}: \\ $$$$ \\ $$$$\mathrm{BD}^{\mathrm{2}} =\mathrm{185}−\mathrm{176cos}\:\mathrm{A}\:\:\:\:\:\mathrm{A}=\left(\mathrm{A1}+\mathrm{A2}\right)\mathrm{O} \\ $$$$\begin{cases}{\mathrm{cos}\:\mathrm{A1}=\frac{\mathrm{11}}{\mathrm{2R}}\:\:\:\:\:\mathrm{sin}\:\mathrm{A1}=\frac{\mathrm{1}}{\mathrm{2R}}\sqrt{\mathrm{4R}^{\mathrm{2}} −\mathrm{121}}}\\{\mathrm{cos}\:\mathrm{A2}=\frac{\mathrm{5}}{\mathrm{R}}\:\:\:\:\:\:\mathrm{sin}\:\mathrm{A2}=\frac{\mathrm{1}}{\mathrm{R}}\sqrt{\mathrm{R}^{\mathrm{2}} −\mathrm{25}}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{149}−\mathrm{140cos}\:\mathrm{C} \\ $$$$ \\ $$$$ \\ $$$$\Rightarrow\:\:\:\mathrm{36}=\mathrm{176cos}\:\mathrm{A}+\mathrm{140cos}\:\mathrm{A} \\ $$$$\Rightarrow\:\:\:\mathrm{36}=\mathrm{316cos}\:\mathrm{A}\:\:\: \\ $$$$\:\:\:\:\:\:\begin{matrix}{}\\{\begin{array}{|c|c|}{\mathrm{cos}\:\mathrm{A}\:\:\:}\\{\:\:\:\frac{\mathrm{9}}{\mathrm{79}}\:\:\:\:\:}\\\hline\end{array}}\end{matrix}\:\:\left[\mathrm{1}\right] \\ $$$$\:\mathrm{sin}\:\mathrm{A}=\mathrm{sinA1cos}\:\mathrm{A2}+\mathrm{sin}\:\mathrm{A2cos}\:\mathrm{A1} \\ $$$$\:\:\mathrm{cos}\:\mathrm{A}=\mathrm{cosA1}\:.\mathrm{cosA2}\:−\mathrm{sinA1}\:.\mathrm{sinA2} \\ $$$$\:\:\: \\ $$$$\:\begin{cases}{}\end{cases} \\ $$
Commented by a.lgnaoui last updated on 26/Jul/24
Answered by A5T last updated on 26/Jul/24
10^2 +11^2 −2(10)(11)cosB=7^2 +8^2 +2(7)(8)cos(B) ⇒cosB=((27)/(83))⇒sinB=((4(√(385)))/(83))⇒AC=((√(1029449))/(83)) ⇒((10×11×((√(1029449))/(83)))/(4R))=(1/2)×10×11×((4(√(385)))/(83)) ⇒R=((√(396337865))/(3089))≈6.4637
$$\mathrm{10}^{\mathrm{2}} +\mathrm{11}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{10}\right)\left(\mathrm{11}\right){cosB}=\mathrm{7}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{7}\right)\left(\mathrm{8}\right){cos}\left({B}\right) \\ $$$$\Rightarrow{cosB}=\frac{\mathrm{27}}{\mathrm{83}}\Rightarrow{sinB}=\frac{\mathrm{4}\sqrt{\mathrm{385}}}{\mathrm{83}}\Rightarrow{AC}=\frac{\sqrt{\mathrm{1029449}}}{\mathrm{83}} \\ $$$$\Rightarrow\frac{\mathrm{10}×\mathrm{11}×\frac{\sqrt{\mathrm{1029449}}}{\mathrm{83}}}{\mathrm{4}{R}}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×\mathrm{11}×\frac{\mathrm{4}\sqrt{\mathrm{385}}}{\mathrm{83}} \\ $$$$\Rightarrow{R}=\frac{\sqrt{\mathrm{396337865}}}{\mathrm{3089}}\approx\mathrm{6}.\mathrm{4637} \\ $$
Answered by mr W last updated on 26/Jul/24
R=(√(((10×11+7×8)(10×8+7×11)(10×7+8×11))/((−10+11+8+7)(10−11+8+7)(10+11−8+7)(10+11+8−7)))) =((√(396337865))/(3080))≈6.4637
$${R}=\sqrt{\frac{\left(\mathrm{10}×\mathrm{11}+\mathrm{7}×\mathrm{8}\right)\left(\mathrm{10}×\mathrm{8}+\mathrm{7}×\mathrm{11}\right)\left(\mathrm{10}×\mathrm{7}+\mathrm{8}×\mathrm{11}\right)}{\left(−\mathrm{10}+\mathrm{11}+\mathrm{8}+\mathrm{7}\right)\left(\mathrm{10}−\mathrm{11}+\mathrm{8}+\mathrm{7}\right)\left(\mathrm{10}+\mathrm{11}−\mathrm{8}+\mathrm{7}\right)\left(\mathrm{10}+\mathrm{11}+\mathrm{8}−\mathrm{7}\right)}} \\ $$$$\:\:\:=\frac{\sqrt{\mathrm{396337865}}}{\mathrm{3080}}\approx\mathrm{6}.\mathrm{4637} \\ $$
Commented by mr W last updated on 26/Jul/24
R=(√(((ab+cd)(ac+bd)(ad+bc))/((−a+b+c+d)(a−b+c+d)(a+b−c+d)(a+b+c−d))))
$${R}=\sqrt{\frac{\left({ab}+{cd}\right)\left({ac}+{bd}\right)\left({ad}+{bc}\right)}{\left(−{a}+{b}+{c}+{d}\right)\left({a}−{b}+{c}+{d}\right)\left({a}+{b}−{c}+{d}\right)\left({a}+{b}+{c}−{d}\right)}} \\ $$

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