Question-219425

Question Number 219425 by Nicholas666 last updated on 24/Apr/25

Commented by Nicholas666 last updated on 24/Apr/25

$$\:\:\:\:{ABC}\:{reguler}\:{pentagon}\: \\ $$$$\:\:\:{P},\:{Q},\:{T}\:\:\:\:\:{toricelli}'{s}\:{point}\:{of}\:{AED},{BCD},{ADE}\:\:\:\:\: \\ $$$$\:\:\:{find}\:{angle}\:{PTQ}? \\ $$

Answered by mr W last updated on 24/Apr/25

Commented by mr W last updated on 24/Apr/25

TB=(2/( (√3))) BQ=((2 cos 36°)/(cos 30°))=((2×2)/( (√3)))×(((√5)+1)/4)=(((√5)+1)/( (√3))) TQ=(√(((2/( (√3))))^2 +((((√5)+1)/( (√3))))^2 −2×(2/( (√3)))×(((√5)+1)/( (√3)))×(((√5)−1)/4))) =(√(((((√5)+1)/( (√3))))^2 ))=(((√5)+1)/( (√3)))=BQ ⇒α=72° x=360°−2×72°−120°=96°

$${TB}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}} \\ $$$${BQ}=\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{36}°}{\mathrm{cos}\:\mathrm{30}°}=\frac{\mathrm{2}×\mathrm{2}}{\:\sqrt{\mathrm{3}}}×\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${TQ}=\sqrt{\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} −\mathrm{2}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}×\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}×\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}} \\ $$$$\:\:\:\:\:=\sqrt{\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}={BQ} \\ $$$$\Rightarrow\alpha=\mathrm{72}° \\ $$$${x}=\mathrm{360}°−\mathrm{2}×\mathrm{72}°−\mathrm{120}°=\mathrm{96}° \\ $$

Commented by Nicholas666 last updated on 25/Apr/25

$${thanks} \\ $$

Answered by Nicholas666 last updated on 29/Apr/25

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