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Question Number 181201 by HeferH last updated on 22/Nov/22

Answered by ARUNG_Brandon_MBU last updated on 23/Nov/22

tanθ=(3/4) tan2θ=(((3/4)+(3/4))/(1−(3/4)∙(3/4)))=((24)/7)=((AC)/(AE))=((AC)/4) ⇒AC=((96)/7) area(ABC)=(1/2)×AB×AC =(1/2)×9×((96)/7)=((432)/7)

$$\mathrm{tan}\theta=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{tan2}\theta=\frac{\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\centerdot\frac{\mathrm{3}}{\mathrm{4}}}=\frac{\mathrm{24}}{\mathrm{7}}=\frac{{AC}}{{AE}}=\frac{{AC}}{\mathrm{4}} \\ $$$$\Rightarrow{AC}=\frac{\mathrm{96}}{\mathrm{7}} \\ $$$$\mathrm{area}\left({ABC}\right)=\frac{\mathrm{1}}{\mathrm{2}}×{AB}×{AC} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{9}×\frac{\mathrm{96}}{\mathrm{7}}=\frac{\mathrm{432}}{\mathrm{7}} \\ $$

Commented by ARUNG_Brandon_MBU last updated on 23/Nov/22

OK

Commented by HeferH last updated on 23/Nov/22

I think that you are asuming AD = 3, which is not necessarily true.

$${I}\:{think}\:{that}\:{you}\:{are}\:{asuming}\:{AD}\:=\:\mathrm{3},\:{which}\: \\ $$$$\:{is}\:{not}\:{necessarily}\:{true}. \\ $$

Answered by mr W last updated on 23/Nov/22

AD=4 tan α AD+DC=4 tan 2α DC=4(tan 2α−tan α)=DB DB=(√(9^2 +(4 tan α)^2 )) 4(tan 2α−tan α)=(√(9^2 +(4 tan α)^2 )) 4 tan α((2/(1−tan^2 α))−1)=(√(81+16 tan^2 α)) 16 tan^2 α(((1+tan^2 α)/(1−tan^2 α)))^2 =81+16 tan^2 α let t=tan^2 α 16t(((1+t)/(1−t)))^2 =81+16t 17t^2 −162t+81=0 t=((81−72)/(17))=(9/(17))=tan^2 α ⇒tan α=(3/( (√(17)))) AC=4×((2×(3/( (√(17)))))/(1−((3/( (√(17)))))^2 ))=3(√(17)) ΔABC=(9/2)×3(√(17))=((27(√(17)))/2) ✓

$${AD}=\mathrm{4}\:\mathrm{tan}\:\alpha \\ $$$${AD}+{DC}=\mathrm{4}\:\mathrm{tan}\:\mathrm{2}\alpha \\ $$$${DC}=\mathrm{4}\left(\mathrm{tan}\:\mathrm{2}\alpha−\mathrm{tan}\:\alpha\right)={DB} \\ $$$${DB}=\sqrt{\mathrm{9}^{\mathrm{2}} +\left(\mathrm{4}\:\mathrm{tan}\:\alpha\right)^{\mathrm{2}} } \\ $$$$\mathrm{4}\left(\mathrm{tan}\:\mathrm{2}\alpha−\mathrm{tan}\:\alpha\right)=\sqrt{\mathrm{9}^{\mathrm{2}} +\left(\mathrm{4}\:\mathrm{tan}\:\alpha\right)^{\mathrm{2}} } \\ $$$$\mathrm{4}\:\mathrm{tan}\:\alpha\left(\frac{\mathrm{2}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\alpha}−\mathrm{1}\right)=\sqrt{\mathrm{81}+\mathrm{16}\:\mathrm{tan}^{\mathrm{2}} \:\alpha} \\ $$$$\mathrm{16}\:\mathrm{tan}^{\mathrm{2}} \:\alpha\left(\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\alpha}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\alpha}\right)^{\mathrm{2}} =\mathrm{81}+\mathrm{16}\:\mathrm{tan}^{\mathrm{2}} \:\alpha \\ $$$${let}\:{t}=\mathrm{tan}^{\mathrm{2}} \:\alpha \\ $$$$\mathrm{16}{t}\left(\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\right)^{\mathrm{2}} =\mathrm{81}+\mathrm{16}{t} \\ $$$$\mathrm{17}{t}^{\mathrm{2}} −\mathrm{162}{t}+\mathrm{81}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{81}−\mathrm{72}}{\mathrm{17}}=\frac{\mathrm{9}}{\mathrm{17}}=\mathrm{tan}^{\mathrm{2}} \:\alpha \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\frac{\mathrm{3}}{\:\sqrt{\mathrm{17}}} \\ $$$${AC}=\mathrm{4}×\frac{\mathrm{2}×\frac{\mathrm{3}}{\:\sqrt{\mathrm{17}}}}{\mathrm{1}−\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{17}}}\right)^{\mathrm{2}} }=\mathrm{3}\sqrt{\mathrm{17}} \\ $$$$\Delta{ABC}=\frac{\mathrm{9}}{\mathrm{2}}×\mathrm{3}\sqrt{\mathrm{17}}=\frac{\mathrm{27}\sqrt{\mathrm{17}}}{\mathrm{2}}\:\checkmark \\ $$

Answered by HeferH last updated on 23/Nov/22

Commented by HeferH last updated on 23/Nov/22

Yes, I used the bisector theorem

$${Yes},\:{I}\:{used}\:{the}\:{bisector}\:{theorem} \\ $$

Commented by HeferH last updated on 23/Nov/22

I′ll share my solution :) thank you all for answering

$$\left.{I}'{ll}\:{share}\:{my}\:{solution}\::\right) \\ $$$$\:{thank}\:{you}\:{all}\:{for}\:{answering}\: \\ $$

Commented by Acem last updated on 23/Nov/22

@Heferh, Sir, your method depends on the circumcircle? or on something else?

$$@{Heferh},\:{Sir},\:{your}\:{method}\:{depends}\:{on} \\ $$$$\:{the}\:{circumcircle}?\:{or}\:{on}\:{something}\:{else}? \\ $$

Commented by HeferH last updated on 23/Nov/22

Commented by HeferH last updated on 23/Nov/22

A cleaner solution (compared to what i did)

$${A}\:{cleaner}\:{solution}\:\left({compared}\:{to}\:{what}\:{i}\:{did}\right) \\ $$$$\: \\ $$

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