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Question Number 11206 by uni last updated on 16/Mar/17

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/Mar/17

AD.AC=6×24=144 AD^2 +AC^2 =24^2 =576 (AD+AC)^2 =576+2×144=6×144⇒ AD+AC=12(√6) t^2 −12(√6)t+144=0⇒ t=((12(√6)±(√(6×144−4×144)))/2)=((12(√6)±12(√2))/2) t_1 =AC=6((√6)+(√2)),t_2 =AD=6((√6)−(√2)) sin C=((AD)/(24))⇒sin C=((6((√6)−(√2)))/(24))⇒C=15^° ∡CDA=75^° ⇒((BD)/(sin15^° ))=(x/(sin(180−75))) BD=x((sin15)/(sin75))=x.tg15 x.AC.sin(90+15)=(BD+24).6⇒ x.6((√6)+(√2))cos15=(xtg15+24)×6 x((√6)+(√2))((((√6)+(√2))/4))=x(2−(√3))+24⇒ x(2+(√3)−2+(√3))=24⇒x=((24)/(2(√3)))=4(√3) ■ t_1 =AC=6((√6)−(√2)),t_2 =AD=6((√6)+(√2)) ⇒∡C=75,∡CDA=15 ((BD)/(sin75))=(x/(sin(180−15)))⇒BD=xcotg15 x.AC.sin(90+15)=(BD+24)×6⇒ x.6((√6)−(√2)).cos15=(x.cotg15+24)×6 x((√6)−(√2))((((√6)+(√2))/4))=x(2+(√3))+24 x(1)=x(2+(√3))+24.......impossible.

$${AD}.{AC}=\mathrm{6}×\mathrm{24}=\mathrm{144} \\ $$$${AD}^{\mathrm{2}} +{AC}^{\mathrm{2}} =\mathrm{24}^{\mathrm{2}} =\mathrm{576} \\ $$$$\left({AD}+{AC}\right)^{\mathrm{2}} =\mathrm{576}+\mathrm{2}×\mathrm{144}=\mathrm{6}×\mathrm{144}\Rightarrow \\ $$$${AD}+{AC}=\mathrm{12}\sqrt{\mathrm{6}} \\ $$$${t}^{\mathrm{2}} −\mathrm{12}\sqrt{\mathrm{6}}{t}+\mathrm{144}=\mathrm{0}\Rightarrow \\ $$$${t}=\frac{\mathrm{12}\sqrt{\mathrm{6}}\pm\sqrt{\mathrm{6}×\mathrm{144}−\mathrm{4}×\mathrm{144}}}{\mathrm{2}}=\frac{\mathrm{12}\sqrt{\mathrm{6}}\pm\mathrm{12}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${t}_{\mathrm{1}} ={AC}=\mathrm{6}\left(\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}\right),{t}_{\mathrm{2}} ={AD}=\mathrm{6}\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right) \\ $$$$\mathrm{sin}\:{C}=\frac{{AD}}{\mathrm{24}}\Rightarrow\mathrm{sin}\:{C}=\frac{\mathrm{6}\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right)}{\mathrm{24}}\Rightarrow{C}=\mathrm{15}^{°} \\ $$$$\measuredangle{CDA}=\mathrm{75}^{°} \Rightarrow\frac{{BD}}{{sin}\mathrm{15}^{°} }=\frac{{x}}{{sin}\left(\mathrm{180}−\mathrm{75}\right)} \\ $$$${BD}={x}\frac{{sin}\mathrm{15}}{{sin}\mathrm{75}}={x}.{tg}\mathrm{15} \\ $$$${x}.{AC}.{sin}\left(\mathrm{90}+\mathrm{15}\right)=\left({BD}+\mathrm{24}\right).\mathrm{6}\Rightarrow \\ $$$${x}.\mathrm{6}\left(\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}\right){cos}\mathrm{15}=\left({xtg}\mathrm{15}+\mathrm{24}\right)×\mathrm{6} \\ $$$${x}\left(\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}\right)\left(\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{4}}\right)={x}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)+\mathrm{24}\Rightarrow \\ $$$${x}\left(\mathrm{2}+\sqrt{\mathrm{3}}−\mathrm{2}+\sqrt{\mathrm{3}}\right)=\mathrm{24}\Rightarrow{x}=\frac{\mathrm{24}}{\mathrm{2}\sqrt{\mathrm{3}}}=\mathrm{4}\sqrt{\mathrm{3}}\:\blacksquare \\ $$$${t}_{\mathrm{1}} ={AC}=\mathrm{6}\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right),{t}_{\mathrm{2}} ={AD}=\mathrm{6}\left(\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}\right) \\ $$$$\Rightarrow\measuredangle{C}=\mathrm{75},\measuredangle{CDA}=\mathrm{15} \\ $$$$\frac{{BD}}{{sin}\mathrm{75}}=\frac{{x}}{{sin}\left(\mathrm{180}−\mathrm{15}\right)}\Rightarrow{BD}={xcotg}\mathrm{15} \\ $$$${x}.{AC}.{sin}\left(\mathrm{90}+\mathrm{15}\right)=\left({BD}+\mathrm{24}\right)×\mathrm{6}\Rightarrow \\ $$$${x}.\mathrm{6}\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right).{cos}\mathrm{15}=\left({x}.{cotg}\mathrm{15}+\mathrm{24}\right)×\mathrm{6} \\ $$$${x}\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right)\left(\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{4}}\right)={x}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)+\mathrm{24} \\ $$$${x}\left(\mathrm{1}\right)={x}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)+\mathrm{24}.......{impossible}. \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 17/Mar/17

DH.CH=6^2 =36,DH+CH=24 CH=((24±(√(24×24−4×36)))/2)=12±6(√3) tgC=((12±6(√3))/6)=2±(√3)⇒∡C=15 ,75 ⇒∡C=15,75⇒∡DAH=15,75(impossible) ⇒∡BAH=30,BH=(1/2)AB=(1/2)x BH=(1/2)x⇒x^2 =((1/2)x)^2 +6^2 ⇒ (3/4)x^2 =36⇒x^2 =48⇒x=4(√3) ■

$${DH}.{CH}=\mathrm{6}^{\mathrm{2}} =\mathrm{36},{DH}+{CH}=\mathrm{24} \\ $$$${CH}=\frac{\mathrm{24}\pm\sqrt{\mathrm{24}×\mathrm{24}−\mathrm{4}×\mathrm{36}}}{\mathrm{2}}=\mathrm{12}\pm\mathrm{6}\sqrt{\mathrm{3}} \\ $$$${tgC}=\frac{\mathrm{12}\pm\mathrm{6}\sqrt{\mathrm{3}}}{\mathrm{6}}=\mathrm{2}\pm\sqrt{\mathrm{3}}\Rightarrow\measuredangle{C}=\mathrm{15}\:,\mathrm{75} \\ $$$$\Rightarrow\measuredangle{C}=\mathrm{15},\mathrm{75}\Rightarrow\measuredangle{DAH}=\mathrm{15},\mathrm{75}\left({impossible}\right) \\ $$$$\Rightarrow\measuredangle{BAH}=\mathrm{30},{BH}=\frac{\mathrm{1}}{\mathrm{2}}{AB}=\frac{\mathrm{1}}{\mathrm{2}}{x} \\ $$$${BH}=\frac{\mathrm{1}}{\mathrm{2}}{x}\Rightarrow{x}^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{2}}{x}\right)^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} \Rightarrow \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}{x}^{\mathrm{2}} =\mathrm{36}\Rightarrow{x}^{\mathrm{2}} =\mathrm{48}\Rightarrow{x}=\mathrm{4}\sqrt{\mathrm{3}}\:\:\:\blacksquare \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 17/Mar/17

x.AC.sin(90+15)=6×(BD+24)..(i) ((BD)/(sin15))=(x/(sin(BDA)))=(x/(sin(90+C)))=(x/(cosC)) AD=24sinC,AC=24cosC AC.AD=6×24⇒24cosC.24sinC=6×24 sin2C=(1/2)⇒2C=30⇒∡C=15 ((BD)/(sin15))=(x/(cos15))⇒BD=xtg15 (i)⇒x.24cos15.cos15=6.(xtg15+24) 4x.cos^2 15=xtg15+24 x(4×((((√6)+(√2))/4))^2 −(2−(√3)))=24⇒ x(2+(√3)−2+(√3))=24⇒x=((24)/(2(√3)))=4(√3) ■

$${x}.{AC}.{sin}\left(\mathrm{90}+\mathrm{15}\right)=\mathrm{6}×\left({BD}+\mathrm{24}\right)..\left({i}\right) \\ $$$$\frac{{BD}}{{sin}\mathrm{15}}=\frac{{x}}{{sin}\left({BDA}\right)}=\frac{{x}}{{sin}\left(\mathrm{90}+{C}\right)}=\frac{{x}}{{cosC}} \\ $$$${AD}=\mathrm{24}{sinC},{AC}=\mathrm{24}{cosC} \\ $$$${AC}.{AD}=\mathrm{6}×\mathrm{24}\Rightarrow\mathrm{24}{cosC}.\mathrm{24}{sinC}=\mathrm{6}×\mathrm{24} \\ $$$${sin}\mathrm{2}{C}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{2}{C}=\mathrm{30}\Rightarrow\measuredangle{C}=\mathrm{15} \\ $$$$\frac{{BD}}{{sin}\mathrm{15}}=\frac{{x}}{{cos}\mathrm{15}}\Rightarrow{BD}={xtg}\mathrm{15} \\ $$$$\left({i}\right)\Rightarrow{x}.\mathrm{24}{cos}\mathrm{15}.{cos}\mathrm{15}=\mathrm{6}.\left({xtg}\mathrm{15}+\mathrm{24}\right) \\ $$$$\mathrm{4}{x}.{cos}^{\mathrm{2}} \mathrm{15}={xtg}\mathrm{15}+\mathrm{24} \\ $$$${x}\left(\mathrm{4}×\left(\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{4}}\right)^{\mathrm{2}} −\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\right)=\mathrm{24}\Rightarrow \\ $$$${x}\left(\mathrm{2}+\sqrt{\mathrm{3}}−\mathrm{2}+\sqrt{\mathrm{3}}\right)=\mathrm{24}\Rightarrow{x}=\frac{\mathrm{24}}{\mathrm{2}\sqrt{\mathrm{3}}}=\mathrm{4}\sqrt{\mathrm{3}}\:\blacksquare \\ $$

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