Question-227373

Question Number 227373 by hardmath last updated on 18/Jan/26

Commented by hardmath last updated on 18/Jan/26

In right-angled triangle ABC ∠A = 90° AB = 8 AC = 6 and ∠B = 2∠BAD Find: CD = ?

$$\mathrm{In}\:\mathrm{right}-\mathrm{angled}\:\mathrm{triangle}\:\mathrm{ABC} \\ $$$$\angle\mathrm{A}\:=\:\mathrm{90}° \\ $$$$\mathrm{AB}\:=\:\mathrm{8} \\ $$$$\mathrm{AC}\:=\:\mathrm{6}\:\:\:\mathrm{and}\:\:\:\angle\mathrm{B}\:=\:\mathrm{2}\angle\mathrm{BAD} \\ $$$$\mathrm{Find}:\:\:\mathrm{CD}\:=\:? \\ $$

Answered by A5T last updated on 19/Jan/26

BD=BC−CD=(√(8^2 +6^2 ))−x=10−x ∠ABF=2α ⇒ ∠BCA=90−2α cos2α=(8/(10))=(4/5)⇒2cos^2 α−1=(4/5)⇒cosα=((√(30))/(10)) ((sinα)/(10−x))=((sin2α)/(AD)) and ((sin(90−α))/x)=((sin(90−2α))/(AD)) ⇒(((10−x)2cosα)/1)=((xcos2α)/(cosα))⇒10−x=((xcos2α)/(2cos^2 α)) 10−x=x×(4/9)⇒x=((90)/(13))

$$\mathrm{BD}=\mathrm{BC}−\mathrm{CD}=\sqrt{\mathrm{8}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} }−\mathrm{x}=\mathrm{10}−\mathrm{x} \\ $$$$\angle\mathrm{ABF}=\mathrm{2}\alpha\:\Rightarrow\:\angle\mathrm{BCA}=\mathrm{90}−\mathrm{2}\alpha \\ $$$$\mathrm{cos2}\alpha=\frac{\mathrm{8}}{\mathrm{10}}=\frac{\mathrm{4}}{\mathrm{5}}\Rightarrow\mathrm{2cos}^{\mathrm{2}} \alpha−\mathrm{1}=\frac{\mathrm{4}}{\mathrm{5}}\Rightarrow\mathrm{cos}\alpha=\frac{\sqrt{\mathrm{30}}}{\mathrm{10}} \\ $$$$\frac{\mathrm{sin}\alpha}{\mathrm{10}−\mathrm{x}}=\frac{\mathrm{sin2}\alpha}{\mathrm{AD}}\:\mathrm{and}\:\frac{\mathrm{sin}\left(\mathrm{90}−\alpha\right)}{\mathrm{x}}=\frac{\mathrm{sin}\left(\mathrm{90}−\mathrm{2}\alpha\right)}{\mathrm{AD}} \\ $$$$\Rightarrow\frac{\left(\mathrm{10}−\mathrm{x}\right)\mathrm{2cos}\alpha}{\mathrm{1}}=\frac{\mathrm{xcos2}\alpha}{\mathrm{cos}\alpha}\Rightarrow\mathrm{10}−\mathrm{x}=\frac{\mathrm{xcos2}\alpha}{\mathrm{2cos}^{\mathrm{2}} \alpha} \\ $$$$\mathrm{10}−\mathrm{x}=\mathrm{x}×\frac{\mathrm{4}}{\mathrm{9}}\Rightarrow\mathrm{x}=\frac{\mathrm{90}}{\mathrm{13}} \\ $$

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