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Question Number 82901 by Power last updated on 25/Feb/20

Answered by mr W last updated on 26/Feb/20

Commented by Power last updated on 26/Feb/20

sir a:b:c=?

$$\mathrm{sir}\:\:\mathrm{a}:\mathrm{b}:\mathrm{c}=? \\ $$

Commented by mr W last updated on 26/Feb/20

let BC=1 a=sin 30=(1/2) ((CF)/(sin 60))=(1/(sin 45)) ⇒CF=((sin 60)/(sin 45))=((√6)/2) AC=(1/(2 cos 75))=(2/((√6)−(√2)))=(((√6)+(√2))/2) AF=(((√6)+(√2))/2)−((√6)/2)=((√2)/2) EF=((√2)/(2(√3)))=((√6)/6) tan α=((√6)/(6×((√6)/2)))=(1/3) HF=b+c=CH=cos 30=((√3)/2) b=HC×tan β=((√3)/2)×tan (45−α) =((√3)/2)×((1−(1/3))/(1+(1/3)))=((√3)/4) c=((√3)/2)−((√3)/4)=((√3)/4) ⇒a=(1/2), b=c=((√3)/4) ⇒a:b:c=2:(√3):(√3)

$${let}\:{BC}=\mathrm{1} \\ $$$${a}=\mathrm{sin}\:\mathrm{30}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{{CF}}{\mathrm{sin}\:\mathrm{60}}=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{45}} \\ $$$$\Rightarrow{CF}=\frac{\mathrm{sin}\:\mathrm{60}}{\mathrm{sin}\:\mathrm{45}}=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$$${AC}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{cos}\:\mathrm{75}}=\frac{\mathrm{2}}{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}=\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${AF}=\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${EF}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\mathrm{3}}}=\frac{\sqrt{\mathrm{6}}}{\mathrm{6}} \\ $$$$\mathrm{tan}\:\alpha=\frac{\sqrt{\mathrm{6}}}{\mathrm{6}×\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${HF}={b}+{c}={CH}=\mathrm{cos}\:\mathrm{30}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${b}={HC}×\mathrm{tan}\:\beta=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\mathrm{tan}\:\left(\mathrm{45}−\alpha\right) \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$${c}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{2}},\:{b}={c}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$\Rightarrow{a}:{b}:{c}=\mathrm{2}:\sqrt{\mathrm{3}}:\sqrt{\mathrm{3}} \\ $$

Commented by Power last updated on 26/Feb/20

thanks

$$\mathrm{thanks} \\ $$

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