Question-220365

Question Number 220365 by fantastic last updated on 11/May/25

Answered by mr W last updated on 11/May/25

Commented by mr W last updated on 11/May/25

α+β=45° tan α=(2/(2x)) tan β=(2/(3x)) ⇒tan β=(2/3)tan α ((tan α+tan β)/(1−tan α tan β))=1 tan α+(2/3) tan α=1−(2/3) tan^2 α 2 tan^2 α+5 tan α−3=0 (2 tan α−1)(tan α+3)=0 ⇒tan α=(1/2) ⇒sin α=(1/( (√5))) sin α=((2−r)/(2+r))=(1/( (√5))) ⇒r=((2((√5)−1))/( (√5)+1))=3−(√5)≈0.764

$$\alpha+\beta=\mathrm{45}° \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{2}}{\mathrm{2}{x}} \\ $$$$\mathrm{tan}\:\beta=\frac{\mathrm{2}}{\mathrm{3}{x}} \\ $$$$\Rightarrow\mathrm{tan}\:\beta=\frac{\mathrm{2}}{\mathrm{3}}\mathrm{tan}\:\alpha \\ $$$$\frac{\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta}{\mathrm{1}−\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta}=\mathrm{1} \\ $$$$\mathrm{tan}\:\alpha+\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{tan}\:\alpha=\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{tan}^{\mathrm{2}} \:\alpha \\ $$$$\mathrm{2}\:\mathrm{tan}^{\mathrm{2}} \:\alpha+\mathrm{5}\:\mathrm{tan}\:\alpha−\mathrm{3}=\mathrm{0} \\ $$$$\left(\mathrm{2}\:\mathrm{tan}\:\alpha−\mathrm{1}\right)\left(\mathrm{tan}\:\alpha+\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{sin}\:\alpha=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$$\mathrm{sin}\:\alpha=\frac{\mathrm{2}−{r}}{\mathrm{2}+{r}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{2}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\:\sqrt{\mathrm{5}}+\mathrm{1}}=\mathrm{3}−\sqrt{\mathrm{5}}\approx\mathrm{0}.\mathrm{764} \\ $$

Commented by fantastic last updated on 11/May/25

$${can}\:{you}\:{do}\:{it}\:{without}\:{trigonometry} \\ $$$${please}\:{do}\:{using}\:{pure}?{geometry} \\ $$

Answered by mr W last updated on 11/May/25

(2x+2)^2 +(3x+2)^2 =(5x)^2 3x^2 −5x−2=0 (3x+1)(x−2)=0 ⇒x=2 (2/(2x))=((2−r)/( (√((2+r)^2 −(2−r)^2 )))) 1=((2−r)/( (√(2r)))) r^2 −6r+4=0 ⇒r=3−(√5) ✓ (r=3+(√5) >2 ⇒rejected)

$$\left(\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{3}{x}+\mathrm{2}\right)^{\mathrm{2}} =\left(\mathrm{5}{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{2}=\mathrm{0} \\ $$$$\left(\mathrm{3}{x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{2} \\ $$$$\frac{\mathrm{2}}{\mathrm{2}{x}}=\frac{\mathrm{2}−{r}}{\:\sqrt{\left(\mathrm{2}+{r}\right)^{\mathrm{2}} −\left(\mathrm{2}−{r}\right)^{\mathrm{2}} }} \\ $$$$\mathrm{1}=\frac{\mathrm{2}−{r}}{\:\sqrt{\mathrm{2}{r}}} \\ $$$${r}^{\mathrm{2}} −\mathrm{6}{r}+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{r}=\mathrm{3}−\sqrt{\mathrm{5}}\:\checkmark \\ $$$$\left({r}=\mathrm{3}+\sqrt{\mathrm{5}}\:>\mathrm{2}\:\Rightarrow{rejected}\right) \\ $$

Commented by mr W last updated on 11/May/25

Answered by cherokeesay last updated on 11/May/25

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