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Question-203327

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Question Number 203327 by mr W last updated on 16/Jan/24
Answered by ajfour last updated on 16/Jan/24
Commented by ajfour last updated on 16/Jan/24
What would r be in this case?
$${What}\:{would}\:{r}\:{be}\:{in}\:{this}\:{case}? \\ $$
Commented by mr W last updated on 16/Jan/24
see below tan φ=(4/6)=(2/3) 2α=45°+φ tan 2α=((1+(2/3))/(1−(2/3)))=5=((2 tan α)/(1−tan^2 α)) 5 tan^2 α+2 tan α−5=0 tan α=(((√(26))−1)/5)=(r/(6−r)) r=((6((√(26))−1))/(4+(√(26))))=3(6−(√(26)))
$${see}\:{below} \\ $$$$\mathrm{tan}\:\phi=\frac{\mathrm{4}}{\mathrm{6}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{2}\alpha=\mathrm{45}°+\phi \\ $$$$\mathrm{tan}\:\mathrm{2}\alpha=\frac{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}}=\mathrm{5}=\frac{\mathrm{2}\:\mathrm{tan}\:\alpha}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\alpha} \\ $$$$\mathrm{5}\:\mathrm{tan}^{\mathrm{2}} \:\alpha+\mathrm{2}\:\mathrm{tan}\:\alpha−\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{tan}\:\alpha=\frac{\sqrt{\mathrm{26}}−\mathrm{1}}{\mathrm{5}}=\frac{{r}}{\mathrm{6}−{r}} \\ $$$${r}=\frac{\mathrm{6}\left(\sqrt{\mathrm{26}}−\mathrm{1}\right)}{\mathrm{4}+\sqrt{\mathrm{26}}}=\mathrm{3}\left(\mathrm{6}−\sqrt{\mathrm{26}}\right) \\ $$
Commented by ajfour last updated on 16/Jan/24
yes sir, excellent! i got the same.
$${yes}\:{sir},\:{excellent}!\:{i}\:{got}\:{the}\:{same}. \\ $$
Commented by mr W last updated on 16/Jan/24
thanks sir!
$${thanks}\:{sir}! \\ $$
Answered by mr W last updated on 16/Jan/24
Commented by mr W last updated on 16/Jan/24
tan φ=(8/(10+6))=(1/2) 2α=45°+φ tan (2α)=((1+(1/2))/(1−(1/2)))=3=((2 tan α)/(1−tan^2 α)) 3 tan^2 α+2 tan α−3=0 ⇒tan α=((−1+(√(10)))/3)=(x/(6−x)) ⇒(2+(√(10)))x=6((√(10))−1) ⇒x=((6((√(10))−1))/(2+(√(10))))=3(4−(√(10))) ✓
$$\mathrm{tan}\:\phi=\frac{\mathrm{8}}{\mathrm{10}+\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}\alpha=\mathrm{45}°+\phi \\ $$$$\mathrm{tan}\:\left(\mathrm{2}\alpha\right)=\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{3}=\frac{\mathrm{2}\:\mathrm{tan}\:\alpha}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\alpha} \\ $$$$\mathrm{3}\:\mathrm{tan}^{\mathrm{2}} \:\alpha+\mathrm{2}\:\mathrm{tan}\:\alpha−\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\frac{−\mathrm{1}+\sqrt{\mathrm{10}}}{\mathrm{3}}=\frac{{x}}{\mathrm{6}−{x}} \\ $$$$\Rightarrow\left(\mathrm{2}+\sqrt{\mathrm{10}}\right){x}=\mathrm{6}\left(\sqrt{\mathrm{10}}−\mathrm{1}\right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{6}\left(\sqrt{\mathrm{10}}−\mathrm{1}\right)}{\mathrm{2}+\sqrt{\mathrm{10}}}=\mathrm{3}\left(\mathrm{4}−\sqrt{\mathrm{10}}\right)\:\checkmark \\ $$

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