Question-207691

Question Number 207691 by cherokeesay last updated on 23/May/24

Answered by mr W last updated on 24/May/24

Commented by mr W last updated on 23/May/24

DE=(√(10^2 −(4+(7/2))^2 ))=((5(√7))/2) sin θ=((5(√7))/(2×10))=((√7)/4) DC=(√(10^2 −(4+(7/2))^2 +((7/2))^2 ))=2(√(14)) R=((2(√(14)))/(2×((√7)/4)))=4(√2)

$${DE}=\sqrt{\mathrm{10}^{\mathrm{2}} −\left(\mathrm{4}+\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\mathrm{5}\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{5}\sqrt{\mathrm{7}}}{\mathrm{2}×\mathrm{10}}=\frac{\sqrt{\mathrm{7}}}{\mathrm{4}} \\ $$$${DC}=\sqrt{\mathrm{10}^{\mathrm{2}} −\left(\mathrm{4}+\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{14}} \\ $$$${R}=\frac{\mathrm{2}\sqrt{\mathrm{14}}}{\mathrm{2}×\frac{\sqrt{\mathrm{7}}}{\mathrm{4}}}=\mathrm{4}\sqrt{\mathrm{2}} \\ $$

Answered by A5T last updated on 23/May/24

(2×10×11−2×4×10)cosθ=10^2 +11^2 −4^2 −10^2 ⇒cosθ=((105)/(140))=(3/4)⇒sinθ=((√7)/4);cos2θ=(1/8);sin2θ=((3(√7))/8) AC=(√(11^2 +4^2 −2×11×4cos2θ))=3(√(14)) ((4×11×3(√(14)))/(4R))=((4×11×((3(√7))/8))/2)⇒((√2)/R)=(1/4)⇒R=4(√2)

$$\left(\mathrm{2}×\mathrm{10}×\mathrm{11}−\mathrm{2}×\mathrm{4}×\mathrm{10}\right){cos}\theta=\mathrm{10}^{\mathrm{2}} +\mathrm{11}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} \\ $$$$\Rightarrow{cos}\theta=\frac{\mathrm{105}}{\mathrm{140}}=\frac{\mathrm{3}}{\mathrm{4}}\Rightarrow{sin}\theta=\frac{\sqrt{\mathrm{7}}}{\mathrm{4}};{cos}\mathrm{2}\theta=\frac{\mathrm{1}}{\mathrm{8}};{sin}\mathrm{2}\theta=\frac{\mathrm{3}\sqrt{\mathrm{7}}}{\mathrm{8}} \\ $$$${AC}=\sqrt{\mathrm{11}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} −\mathrm{2}×\mathrm{11}×\mathrm{4}{cos}\mathrm{2}\theta}=\mathrm{3}\sqrt{\mathrm{14}} \\ $$$$\frac{\mathrm{4}×\mathrm{11}×\mathrm{3}\sqrt{\mathrm{14}}}{\mathrm{4}{R}}=\frac{\mathrm{4}×\mathrm{11}×\frac{\mathrm{3}\sqrt{\mathrm{7}}}{\mathrm{8}}}{\mathrm{2}}\Rightarrow\frac{\sqrt{\mathrm{2}}}{{R}}=\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow{R}=\mathrm{4}\sqrt{\mathrm{2}} \\ $$

Commented by cherokeesay last updated on 23/May/24

$${nice}\:!\:{thank}\:{you}. \\ $$

Answered by mr W last updated on 23/May/24

AD^2 =4^2 +10^2 −2×4×10 cos θ DC^2 =11^2 +10^2 −2×11×10 cos θ AD=DC 4^2 +10^2 −2×4×10 cos θ=11^2 +10^2 −2×11×10 cos θ 2(11−4)×10 cos θ=11^2 −4^2 cos θ=((11+4)/(2×10))=(3/4) ⇒sin θ=((√7)/4) DC=(√(11^2 +10^2 −2×11×10×(3/4)))=2(√(14)) R=((DC)/(2 sin θ))=((2(√(14)))/(2×((√7)/4)))=4(√2) ✓

$${AD}^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} −\mathrm{2}×\mathrm{4}×\mathrm{10}\:\mathrm{cos}\:\theta \\ $$$${DC}^{\mathrm{2}} =\mathrm{11}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} −\mathrm{2}×\mathrm{11}×\mathrm{10}\:\mathrm{cos}\:\theta \\ $$$${AD}={DC} \\ $$$$\mathrm{4}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} −\mathrm{2}×\mathrm{4}×\mathrm{10}\:\mathrm{cos}\:\theta=\mathrm{11}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} −\mathrm{2}×\mathrm{11}×\mathrm{10}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{2}\left(\mathrm{11}−\mathrm{4}\right)×\mathrm{10}\:\mathrm{cos}\:\theta=\mathrm{11}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{11}+\mathrm{4}}{\mathrm{2}×\mathrm{10}}=\frac{\mathrm{3}}{\mathrm{4}}\:\Rightarrow\mathrm{sin}\:\theta=\frac{\sqrt{\mathrm{7}}}{\mathrm{4}} \\ $$$${DC}=\sqrt{\mathrm{11}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} −\mathrm{2}×\mathrm{11}×\mathrm{10}×\frac{\mathrm{3}}{\mathrm{4}}}=\mathrm{2}\sqrt{\mathrm{14}} \\ $$$${R}=\frac{{DC}}{\mathrm{2}\:\mathrm{sin}\:\theta}=\frac{\mathrm{2}\sqrt{\mathrm{14}}}{\mathrm{2}×\frac{\sqrt{\mathrm{7}}}{\mathrm{4}}}=\mathrm{4}\sqrt{\mathrm{2}}\:\checkmark \\ $$

Commented by cherokeesay last updated on 23/May/24

$${nice}\:{thank}\:{you}\:{master}. \\ $$

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