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Question-81188

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Question Number 81188 by jagoll last updated on 10/Feb/20
Commented by jagoll last updated on 10/Feb/20
dear mr W
$${dear}\:{mr}\:{W} \\ $$
Answered by mr W last updated on 10/Feb/20
Commented by mr W last updated on 10/Feb/20
WAY 1 R_A =2+1=3 AB=3−2=1 AC=3−1=2 BD=2+R_D AD=3−R_D CD=1+R_D cos α=((2^2 +(3−R_D )^2 −(1+R_D )^2 )/(2×2(3−R_D ))) cos β=((1^2 +(3−R_D )^2 −(2+R_D )^2 )/(2(3−R_D ))) α+β=180° cos α=−cos β ((2^2 +(3−R_D )^2 −(1+R_D )^2 )/(2×2(3−R_D )))=−((1^2 +(3−R_D )^2 −(2+R_D )^2 )/(2(3−R_D ))) ((4+4(2−2R_D ))/2)=−((1+5(1−2R_D ))/1) 6=7R_D ⇒R_D =(6/7)
$${WAY}\:\:\mathrm{1} \\ $$$${R}_{{A}} =\mathrm{2}+\mathrm{1}=\mathrm{3} \\ $$$${AB}=\mathrm{3}−\mathrm{2}=\mathrm{1} \\ $$$${AC}=\mathrm{3}−\mathrm{1}=\mathrm{2} \\ $$$${BD}=\mathrm{2}+{R}_{{D}} \\ $$$${AD}=\mathrm{3}−{R}_{{D}} \\ $$$${CD}=\mathrm{1}+{R}_{{D}} \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{2}^{\mathrm{2}} +\left(\mathrm{3}−{R}_{{D}} \right)^{\mathrm{2}} −\left(\mathrm{1}+{R}_{{D}} \right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}\left(\mathrm{3}−{R}_{{D}} \right)} \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{1}^{\mathrm{2}} +\left(\mathrm{3}−{R}_{{D}} \right)^{\mathrm{2}} −\left(\mathrm{2}+{R}_{{D}} \right)^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{3}−{R}_{{D}} \right)} \\ $$$$\alpha+\beta=\mathrm{180}° \\ $$$$\mathrm{cos}\:\alpha=−\mathrm{cos}\:\beta \\ $$$$\frac{\mathrm{2}^{\mathrm{2}} +\left(\mathrm{3}−{R}_{{D}} \right)^{\mathrm{2}} −\left(\mathrm{1}+{R}_{{D}} \right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}\left(\mathrm{3}−{R}_{{D}} \right)}=−\frac{\mathrm{1}^{\mathrm{2}} +\left(\mathrm{3}−{R}_{{D}} \right)^{\mathrm{2}} −\left(\mathrm{2}+{R}_{{D}} \right)^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{3}−{R}_{{D}} \right)} \\ $$$$\frac{\mathrm{4}+\mathrm{4}\left(\mathrm{2}−\mathrm{2}{R}_{{D}} \right)}{\mathrm{2}}=−\frac{\mathrm{1}+\mathrm{5}\left(\mathrm{1}−\mathrm{2}{R}_{{D}} \right)}{\mathrm{1}} \\ $$$$\mathrm{6}=\mathrm{7}{R}_{{D}} \\ $$$$\Rightarrow{R}_{{D}} =\frac{\mathrm{6}}{\mathrm{7}} \\ $$
Commented by mr W last updated on 10/Feb/20
WAY 2 (−(1/3)+(1/2)+(1/1)+(1/R_D ))^2 =2((1/3^2 )+(1/2^2 )+(1/1^2 )+(1/R_D ^2 )) (1/R_D ^2 )−(7/(3R_D ))+((49)/(36))=0 ((1/R_D )−(7/6))^2 =0 ⇒R_D =(6/7)
$${WAY}\:\:\mathrm{2} \\ $$$$\left(−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{{R}_{{D}} }\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{{R}_{{D}} ^{\mathrm{2}} }\right) \\ $$$$\frac{\mathrm{1}}{{R}_{{D}} ^{\mathrm{2}} }−\frac{\mathrm{7}}{\mathrm{3}{R}_{{D}} }+\frac{\mathrm{49}}{\mathrm{36}}=\mathrm{0} \\ $$$$\left(\frac{\mathrm{1}}{{R}_{{D}} }−\frac{\mathrm{7}}{\mathrm{6}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{R}_{{D}} =\frac{\mathrm{6}}{\mathrm{7}} \\ $$
Commented by jagoll last updated on 10/Feb/20
thank you mister. i like way 1
$${thank}\:{you}\:{mister}.\:{i}\:{like}\:{way}\:\mathrm{1} \\ $$
Answered by ajfour last updated on 10/Feb/20
(x−h)^2 +(y−k)^2 =r^2 h^2 +k^2 =(3−r)^2 (h+1)^2 +k^2 =(2+r)^2 (2−h)^2 +k^2 =(1+r)^2 ⇒ (3−r)^2 +2h+1=(2+r)^2 & 4−4h+(3−r)^2 =(1+r)^2 ⇒ 6+3(3−r)^2 =2(2+r)^2 +(1+r)^2 ⇒ 28r=24 ⇒ r=(6/7) .
$$\left({x}−{h}\right)^{\mathrm{2}} +\left({y}−{k}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${h}^{\mathrm{2}} +{k}^{\mathrm{2}} =\left(\mathrm{3}−{r}\right)^{\mathrm{2}} \\ $$$$\left({h}+\mathrm{1}\right)^{\mathrm{2}} +{k}^{\mathrm{2}} =\left(\mathrm{2}+{r}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{2}−{h}\right)^{\mathrm{2}} +{k}^{\mathrm{2}} =\left(\mathrm{1}+{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\left(\mathrm{3}−{r}\right)^{\mathrm{2}} +\mathrm{2}{h}+\mathrm{1}=\left(\mathrm{2}+{r}\right)^{\mathrm{2}} \\ $$$$\&\:\:\:\mathrm{4}−\mathrm{4}{h}+\left(\mathrm{3}−{r}\right)^{\mathrm{2}} =\left(\mathrm{1}+{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{6}+\mathrm{3}\left(\mathrm{3}−{r}\right)^{\mathrm{2}} =\mathrm{2}\left(\mathrm{2}+{r}\right)^{\mathrm{2}} +\left(\mathrm{1}+{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{28}{r}=\mathrm{24}\:\:\:\Rightarrow\:\:\boldsymbol{{r}}=\frac{\mathrm{6}}{\mathrm{7}}\:. \\ $$
Commented by jagoll last updated on 10/Feb/20
thank you mister . great..
$${thank}\:{you}\:{mister}\:.\:{great}.. \\ $$

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