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Question-59342




Question Number 59342 by naka3546 last updated on 08/May/19
Answered by mr W last updated on 08/May/19
Commented by mr W last updated on 08/May/19
let radius of circle=1  ⇒side length of square=(1/( (√2)))  ((OB)/(sin ∠OFB))=((OF)/(sin ∠OBF))  ⇒sin ∠OFB=(1/( (√2)))×(1/( (√2)))=(1/2)  ⇒∠OFB=30°  ∠OFG=∠OGF=40−30=10°  ∠FOD=45−30=15°  ∠FOE=45+15=60°  ∠FGE=60/2=30°  ∠OGE=∠OEG=30−10=20°  ⇒x=45−20=25°
$${let}\:{radius}\:{of}\:{circle}=\mathrm{1} \\ $$$$\Rightarrow{side}\:{length}\:{of}\:{square}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\frac{{OB}}{\mathrm{sin}\:\angle{OFB}}=\frac{{OF}}{\mathrm{sin}\:\angle{OBF}} \\ $$$$\Rightarrow\mathrm{sin}\:\angle{OFB}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\angle{OFB}=\mathrm{30}° \\ $$$$\angle{OFG}=\angle{OGF}=\mathrm{40}−\mathrm{30}=\mathrm{10}° \\ $$$$\angle{FOD}=\mathrm{45}−\mathrm{30}=\mathrm{15}° \\ $$$$\angle{FOE}=\mathrm{45}+\mathrm{15}=\mathrm{60}° \\ $$$$\angle{FGE}=\mathrm{60}/\mathrm{2}=\mathrm{30}° \\ $$$$\angle{OGE}=\angle{OEG}=\mathrm{30}−\mathrm{10}=\mathrm{20}° \\ $$$$\Rightarrow{x}=\mathrm{45}−\mathrm{20}=\mathrm{25}° \\ $$

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