Question-204851

Question Number 204851 by ibrahimabdullayev last updated on 28/Feb/24

Answered by A5T last updated on 29/Feb/24

5^2 =6^2 +5^2 −2×6×5cos∠ACB⇒cos∠ACB=(3/5) 6^2 =5^2 +5^2 −2×5×5cos∠ABC⇒cos∠ABC=(7/(25)) sin(∠ACB−a)=((BP)/5) =sin∠ACBcosa−cos∠ACBsina...(i) 180−(90−∠BCA+a+2∠BCA)=∠ABP =90−a−∠BCA=∠ABP ((sina)/(BP))=((sin(90+∠BCA))/5)=((cos(−BCA))/5)=(3/(25)) ⇒sina=((3BP)/(25))...(ii) (ii)⇒(i): ((BP)/5)=((4cosa)/5)−((3sina)/5)⇒BP+((9BP)/(125))=4cosa cosa=((134BP)/(500)) ((sina)/(BD))=((sin(180−a−∠ABD))/5)=((cos(−90+a+∠ABD))/5) =((cos(a−90)cos(ABD)−sin(a−90)sin(ABD))/5) =((((21BP)/(625))+((134BP)/(500))×((24)/(25)))/5)=(((909BP)/(3125))/5)=((909BP)/(15625)) ⇒BD=((15625sina)/(909BP))=((15625×3)/(909×25))=((625)/(303)) BD+DC=5⇒DC=5−((625)/(303))=((890)/(303)) ⇒BD:DC=625:890=125:178

$$\mathrm{5}^{\mathrm{2}} =\mathrm{6}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{2}×\mathrm{6}×\mathrm{5}{cos}\angle{ACB}\Rightarrow{cos}\angle{ACB}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\mathrm{6}^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{2}×\mathrm{5}×\mathrm{5}{cos}\angle{ABC}\Rightarrow{cos}\angle{ABC}=\frac{\mathrm{7}}{\mathrm{25}} \\ $$$${sin}\left(\angle{ACB}−{a}\right)=\frac{{BP}}{\mathrm{5}} \\ $$$$={sin}\angle{ACBcosa}−{cos}\angle{ACBsina}…\left({i}\right) \\ $$$$\mathrm{180}−\left(\mathrm{90}−\angle{BCA}+{a}+\mathrm{2}\angle{BCA}\right)=\angle{ABP} \\ $$$$=\mathrm{90}−{a}−\angle{BCA}=\angle{ABP} \\ $$$$\frac{{sina}}{{BP}}=\frac{{sin}\left(\mathrm{90}+\angle{BCA}\right)}{\mathrm{5}}=\frac{{cos}\left(−{BCA}\right)}{\mathrm{5}}=\frac{\mathrm{3}}{\mathrm{25}} \\ $$$$\Rightarrow{sina}=\frac{\mathrm{3}{BP}}{\mathrm{25}}…\left({ii}\right) \\ $$$$\left({ii}\right)\Rightarrow\left({i}\right):\:\frac{{BP}}{\mathrm{5}}=\frac{\mathrm{4}{cosa}}{\mathrm{5}}−\frac{\mathrm{3}{sina}}{\mathrm{5}}\Rightarrow{BP}+\frac{\mathrm{9}{BP}}{\mathrm{125}}=\mathrm{4}{cosa} \\ $$$${cosa}=\frac{\mathrm{134}{BP}}{\mathrm{500}} \\ $$$$\frac{{sina}}{{BD}}=\frac{{sin}\left(\mathrm{180}−{a}−\angle{ABD}\right)}{\mathrm{5}}=\frac{{cos}\left(−\mathrm{90}+{a}+\angle{ABD}\right)}{\mathrm{5}} \\ $$$$=\frac{{cos}\left({a}−\mathrm{90}\right){cos}\left({ABD}\right)−{sin}\left({a}−\mathrm{90}\right){sin}\left({ABD}\right)}{\mathrm{5}} \\ $$$$=\frac{\frac{\mathrm{21}{BP}}{\mathrm{625}}+\frac{\mathrm{134}{BP}}{\mathrm{500}}×\frac{\mathrm{24}}{\mathrm{25}}}{\mathrm{5}}=\frac{\frac{\mathrm{909}{BP}}{\mathrm{3125}}}{\mathrm{5}}=\frac{\mathrm{909}{BP}}{\mathrm{15625}} \\ $$$$\Rightarrow{BD}=\frac{\mathrm{15625}{sina}}{\mathrm{909}{BP}}=\frac{\mathrm{15625}×\mathrm{3}}{\mathrm{909}×\mathrm{25}}=\frac{\mathrm{625}}{\mathrm{303}} \\ $$$${BD}+{DC}=\mathrm{5}\Rightarrow{DC}=\mathrm{5}−\frac{\mathrm{625}}{\mathrm{303}}=\frac{\mathrm{890}}{\mathrm{303}} \\ $$$$\Rightarrow{BD}:{DC}=\mathrm{625}:\mathrm{890}=\mathrm{125}:\mathrm{178} \\ $$

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