Question Number 203602 by mr W last updated on 22/Jan/24
Answered by mr W last updated on 23/Jan/24
Commented by mr W last updated on 23/Jan/24
$${make}\:{PE}//{CD} \\ $$$$\angle{EPD}=\angle{PDC}=\mathrm{60}°=\angle{PDE} \\ $$$$\Rightarrow\Delta{PDE}\:{is}\:{equilateral}. \\ $$$$\Rightarrow{PE}={ED}={PD} \\ $$$$\frac{{BE}}{{BD}}=\frac{{EP}}{{DC}} \\ $$$$\frac{{BD}−{ED}}{{BD}}=\frac{{EP}}{{DC}} \\ $$$$\frac{\mathrm{24}−{PD}}{\mathrm{24}}=\frac{{PD}}{\mathrm{8}} \\ $$$$\mathrm{4}×{PD}=\mathrm{24} \\ $$$$\Rightarrow{PD}=\mathrm{6}\:\checkmark \\ $$
Answered by AST last updated on 22/Jan/24
![x=AB=BC=CA Ptolemy: 8×x+24×x=x(DA)⇒DA=32 Let ∡CBD=θ⇒∠BCD=60−θ BC^2 =BD^2 +DC^2 −2BD×DCcos(120) ⇒BC=8(√(13)) ((sin(θ))/8)=((sin(120))/(8(√(13))))⇒sin(θ)=((√(39))/(26))⇒cos(θ)=((7(√(13)))/(26)) ((sin(120−θ))/(24))=((sin(θ))/(DP))⇒DP=((24sin(θ))/(sin(120−θ)))=6 [sin(120−θ)=((8(√(39)))/(52))]](https://www.tinkutara.com/question/Q203608.png)
$${x}={AB}={BC}={CA} \\ $$$${Ptolemy}:\:\mathrm{8}×{x}+\mathrm{24}×{x}={x}\left({DA}\right)\Rightarrow{DA}=\mathrm{32} \\ $$$${Let}\:\measuredangle{CBD}=\theta\Rightarrow\angle{BCD}=\mathrm{60}−\theta \\ $$$${BC}^{\mathrm{2}} ={BD}^{\mathrm{2}} +{DC}^{\mathrm{2}} −\mathrm{2}{BD}×{DCcos}\left(\mathrm{120}\right) \\ $$$$\Rightarrow{BC}=\mathrm{8}\sqrt{\mathrm{13}} \\ $$$$\frac{{sin}\left(\theta\right)}{\mathrm{8}}=\frac{{sin}\left(\mathrm{120}\right)}{\mathrm{8}\sqrt{\mathrm{13}}}\Rightarrow{sin}\left(\theta\right)=\frac{\sqrt{\mathrm{39}}}{\mathrm{26}}\Rightarrow{cos}\left(\theta\right)=\frac{\mathrm{7}\sqrt{\mathrm{13}}}{\mathrm{26}} \\ $$$$\frac{{sin}\left(\mathrm{120}−\theta\right)}{\mathrm{24}}=\frac{{sin}\left(\theta\right)}{{DP}}\Rightarrow{DP}=\frac{\mathrm{24}{sin}\left(\theta\right)}{{sin}\left(\mathrm{120}−\theta\right)}=\mathrm{6} \\ $$$$\left[{sin}\left(\mathrm{120}−\theta\right)=\frac{\mathrm{8}\sqrt{\mathrm{39}}}{\mathrm{52}}\right] \\ $$
Commented by mr W last updated on 23/Jan/24
$${thanks}\:{alot}! \\ $$