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Question-56986




Question Number 56986 by mr W last updated on 27/Mar/19
Commented by mr W last updated on 27/Mar/19
Find x=?
$${Find}\:{x}=? \\ $$
Answered by MJS last updated on 28/Mar/19
Commented by mr W last updated on 28/Mar/19
correct, thank you sir!
$${correct},\:{thank}\:{you}\:{sir}! \\ $$
Answered by mr W last updated on 28/Mar/19
Commented by mr W last updated on 28/Mar/19
alternative way:  AB=2×5=10  AC=(√(10^2 −6^2 ))=8  cos (2α)=((AC)/(AB))=(8/(10))=(4/5)  2 cos^2  α−1=(4/5)  ⇒cos α=(3/( (√(10))))  AD=AC cos α=((24)/( (√(10))))  FD=DC=AC sin α=8(√(1−(9/(10))))=(8/( (√(10))))  ((ED)/(AD))=((BC)/(AC))=(3/4)  ⇒ED=(3/4)×((24)/( (√(10))))=((18)/( (√(10))))  x=ED−FD=((18)/( (√(10))))−(8/( (√(10))))=((10)/( (√(10))))=(√(10))
$${alternative}\:{way}: \\ $$$${AB}=\mathrm{2}×\mathrm{5}=\mathrm{10} \\ $$$${AC}=\sqrt{\mathrm{10}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} }=\mathrm{8} \\ $$$$\mathrm{cos}\:\left(\mathrm{2}\alpha\right)=\frac{{AC}}{{AB}}=\frac{\mathrm{8}}{\mathrm{10}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\alpha−\mathrm{1}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{cos}\:\alpha=\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}} \\ $$$${AD}={AC}\:\mathrm{cos}\:\alpha=\frac{\mathrm{24}}{\:\sqrt{\mathrm{10}}} \\ $$$${FD}={DC}={AC}\:\mathrm{sin}\:\alpha=\mathrm{8}\sqrt{\mathrm{1}−\frac{\mathrm{9}}{\mathrm{10}}}=\frac{\mathrm{8}}{\:\sqrt{\mathrm{10}}} \\ $$$$\frac{{ED}}{{AD}}=\frac{{BC}}{{AC}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{ED}=\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{24}}{\:\sqrt{\mathrm{10}}}=\frac{\mathrm{18}}{\:\sqrt{\mathrm{10}}} \\ $$$${x}={ED}−{FD}=\frac{\mathrm{18}}{\:\sqrt{\mathrm{10}}}−\frac{\mathrm{8}}{\:\sqrt{\mathrm{10}}}=\frac{\mathrm{10}}{\:\sqrt{\mathrm{10}}}=\sqrt{\mathrm{10}} \\ $$

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