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Question-76087




Question Number 76087 by aliesam last updated on 23/Dec/19
Answered by mr W last updated on 23/Dec/19
Commented by mr W last updated on 23/Dec/19
AB=(√(2^2 +6^2 ))=2(√(10))  AC=(√(10))  AO=R=(√(50))  cos ∠OAC=((√(10))/( (√(50))))=(1/( (√5)))  cos ∠DAC=(6/(2(√(10))))=(3/( (√(10))))  cos α=cos (∠OAC−∠DAC)  =(1/( (√5)))×(3/( (√(10))))+(2/( (√5)))×(1/( (√(10))))=((√2)/2)  x^2 =OD^2 =((√(50)))^2 +6^2 −2(√(50))×6×cos α  =50+36−2(√(50))×6×((√2)/2)=26  ⇒x=(√(26))
$${AB}=\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{10}} \\ $$$${AC}=\sqrt{\mathrm{10}} \\ $$$${AO}={R}=\sqrt{\mathrm{50}} \\ $$$$\mathrm{cos}\:\angle{OAC}=\frac{\sqrt{\mathrm{10}}}{\:\sqrt{\mathrm{50}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$$\mathrm{cos}\:\angle{DAC}=\frac{\mathrm{6}}{\mathrm{2}\sqrt{\mathrm{10}}}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}} \\ $$$$\mathrm{cos}\:\alpha=\mathrm{cos}\:\left(\angle{OAC}−\angle{DAC}\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}×\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} ={OD}^{\mathrm{2}} =\left(\sqrt{\mathrm{50}}\right)^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{50}}×\mathrm{6}×\mathrm{cos}\:\alpha \\ $$$$=\mathrm{50}+\mathrm{36}−\mathrm{2}\sqrt{\mathrm{50}}×\mathrm{6}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=\mathrm{26} \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{26}} \\ $$

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