Question-199730

Question Number 199730 by Rupesh123 last updated on 08/Nov/23

Answered by mr W last updated on 10/Nov/23

Commented by mr W last updated on 10/Nov/23

(x/(sin θ))=(1/(sin 18°))=(y/(sin (θ−18°))) ((y+1)/(sin (θ+18°)))=(1/(sin 18°)) y=((sin (θ−18°))/(sin 18°))=((sin (θ+18°))/(sin 18°))−1 ((sin (θ+18°)−sin (θ−18°))/(sin 18°))=1 ⇒cos θ=(1/2) ⇒sin θ=((√3)/2) x=((sin θ)/(sin 18°))=((√3)/(2 sin 18°))=((4(√3))/(2((√5)−1))) =(((√(15))+(√3))/2) ✓

$$\frac{{x}}{\mathrm{sin}\:\theta}=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{18}°}=\frac{{y}}{\mathrm{sin}\:\left(\theta−\mathrm{18}°\right)} \\ $$$$\frac{{y}+\mathrm{1}}{\mathrm{sin}\:\left(\theta+\mathrm{18}°\right)}=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{18}°} \\ $$$${y}=\frac{\mathrm{sin}\:\left(\theta−\mathrm{18}°\right)}{\mathrm{sin}\:\mathrm{18}°}=\frac{\mathrm{sin}\:\left(\theta+\mathrm{18}°\right)}{\mathrm{sin}\:\mathrm{18}°}−\mathrm{1} \\ $$$$\frac{\mathrm{sin}\:\left(\theta+\mathrm{18}°\right)−\mathrm{sin}\:\left(\theta−\mathrm{18}°\right)}{\mathrm{sin}\:\mathrm{18}°}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{sin}\:\theta=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${x}=\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\mathrm{18}°}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{18}°}=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{2}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)} \\ $$$$\:\:=\frac{\sqrt{\mathrm{15}}+\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\checkmark \\ $$

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Question-199730

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