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Question-177047




Question Number 177047 by Ar Brandon last updated on 30/Sep/22
Answered by som(math1967) last updated on 30/Sep/22
cosD=((7^2 +DC^2 −5^2 )/(2×7DC))=((24+DC^2 )/(14DC))  again  cosD=((7^2 +DB^2 −5^2 )/(2×7×DB))=((24+DB^2 )/(14DB))  ((24+DC^2 )/(14DC))=((24+DB^2 )/(14DB))  ⇒24DB+DB×DC^2 =24DC+DC×DB^2   ⇒24(DB−DC)=DB×DC(DB−DC)  DB×DC=24 [∵DB≠DC]
$${cosD}=\frac{\mathrm{7}^{\mathrm{2}} +{DC}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} }{\mathrm{2}×\mathrm{7}{DC}}=\frac{\mathrm{24}+{DC}^{\mathrm{2}} }{\mathrm{14}{DC}} \\ $$$${again}\:\:{cosD}=\frac{\mathrm{7}^{\mathrm{2}} +{DB}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} }{\mathrm{2}×\mathrm{7}×{DB}}=\frac{\mathrm{24}+{DB}^{\mathrm{2}} }{\mathrm{14}{DB}} \\ $$$$\frac{\mathrm{24}+{DC}^{\mathrm{2}} }{\mathrm{14}{DC}}=\frac{\mathrm{24}+{DB}^{\mathrm{2}} }{\mathrm{14}{DB}} \\ $$$$\Rightarrow\mathrm{24}{DB}+{DB}×{DC}^{\mathrm{2}} =\mathrm{24}{DC}+{DC}×{DB}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{24}\left({DB}−{DC}\right)={DB}×{DC}\left({DB}−{DC}\right) \\ $$$${DB}×{DC}=\mathrm{24}\:\left[\because{DB}\neq{DC}\right] \\ $$$$ \\ $$
Commented by Ar Brandon last updated on 30/Sep/22
Thank you Sir
Answered by HeferH last updated on 30/Sep/22
    Let h be the height of △ABC    h^2  = 25 − ((BC^2 )/4)   h^2  + (((BC)/2) + CD)^2  = 49   CD^2  + (CD)(BC)=24   CD(CD + BC) = 24   CD(DB)=24
$$\: \\ $$$$\:{Let}\:{h}\:{be}\:{the}\:{height}\:{of}\:\bigtriangleup{ABC}\: \\ $$$$\:{h}^{\mathrm{2}} \:=\:\mathrm{25}\:−\:\frac{{BC}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:{h}^{\mathrm{2}} \:+\:\left(\frac{{BC}}{\mathrm{2}}\:+\:{CD}\right)^{\mathrm{2}} \:=\:\mathrm{49} \\ $$$$\:{CD}^{\mathrm{2}} \:+\:\left({CD}\right)\left({BC}\right)=\mathrm{24} \\ $$$$\:{CD}\left({CD}\:+\:{BC}\right)\:=\:\mathrm{24} \\ $$$$\:{CD}\left({DB}\right)=\mathrm{24} \\ $$
Commented by Ar Brandon last updated on 30/Sep/22
Thank you Sir

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