Question and Answers Forum

All Questions      Topic List

Vector Questions

Previous in All Question      Next in All Question      

Previous in Vector      Next in Vector      

Question Number 88214 by M±th+et£s last updated on 09/Apr/20

Answered by ajfour last updated on 09/Apr/20

AB=t , radius=r , AD=DC=b upper section of OB=c t^2 =2b^2 ; tan α=(r/(t+r)) ((√(r^2 +c^2 ))/(2b))=(c/r)=(r/(t+r)) ⇒ ((√(r^2 +c^2 ))/(t(√2)))=(c/r)=(r/(t+r)) ⇒ c=(r^2 /(t+r)) ⇒ ((r^2 +((r^2 /(t+r)))^2 )/(2t^2 ))=(r^2 /((t+r)^2 )) let t/r=s ⇒ ((1+(1/((s+1)^2 )))/(2s^2 ))=(1/((s+1)^2 )) ⇒ (s+1)^2 +1=2s^2 s^2 −2s−2=0 s=1+(√3) tan α=(1/(s+1)) = (1/(2+(√3))) = 2−(√3) . ⇒ 𝛂=tan^(−1) (2−(√3) )=15° .

$${AB}={t}\:\:,\:\:{radius}={r}\:,\:{AD}={DC}={b} \\ $$$${upper}\:{section}\:{of}\:{OB}={c} \\ $$$${t}^{\mathrm{2}} =\mathrm{2}{b}^{\mathrm{2}} \:\:\:;\:\:\:\mathrm{tan}\:\alpha=\frac{{r}}{{t}+{r}} \\ $$$$\frac{\sqrt{{r}^{\mathrm{2}} +{c}^{\mathrm{2}} }}{\mathrm{2}{b}}=\frac{{c}}{{r}}=\frac{{r}}{{t}+{r}} \\ $$$$\Rightarrow\:\:\:\frac{\sqrt{{r}^{\mathrm{2}} +{c}^{\mathrm{2}} }}{{t}\sqrt{\mathrm{2}}}=\frac{{c}}{{r}}=\frac{{r}}{{t}+{r}} \\ $$$$\:\:\Rightarrow\:\:{c}=\frac{{r}^{\mathrm{2}} }{{t}+{r}} \\ $$$$\Rightarrow\:\:\frac{{r}^{\mathrm{2}} +\left(\frac{{r}^{\mathrm{2}} }{{t}+{r}}\right)^{\mathrm{2}} }{\mathrm{2}{t}^{\mathrm{2}} }=\frac{{r}^{\mathrm{2}} }{\left({t}+{r}\right)^{\mathrm{2}} } \\ $$$${let}\:\:{t}/{r}={s} \\ $$$$\Rightarrow\:\:\:\:\frac{\mathrm{1}+\frac{\mathrm{1}}{\left({s}+\mathrm{1}\right)^{\mathrm{2}} }}{\mathrm{2}{s}^{\mathrm{2}} }=\frac{\mathrm{1}}{\left({s}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\left({s}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}=\mathrm{2}{s}^{\mathrm{2}} \\ $$$${s}^{\mathrm{2}} −\mathrm{2}{s}−\mathrm{2}=\mathrm{0} \\ $$$${s}=\mathrm{1}+\sqrt{\mathrm{3}} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{{s}+\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}}\:=\:\mathrm{2}−\sqrt{\mathrm{3}}\:. \\ $$$$\:\:\Rightarrow\:\:\boldsymbol{\alpha}=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}−\sqrt{\mathrm{3}}\:\right)=\mathrm{15}°\:. \\ $$

Commented by mr W last updated on 09/Apr/20

correct!

$${correct}! \\ $$

Commented by M±th+et£s last updated on 09/Apr/20

nice solution sir

$${nice}\:{solution}\:{sir} \\ $$

Answered by mr W last updated on 09/Apr/20

Commented by mr W last updated on 09/Apr/20

r=radius AD=DC ⇒CF=((OB)/2)=(r/2) CD=CE×cos α=2r cos α CF=CD×sin α=2r cos α sin α=r sin 2α ⇒r sin 2α=(r/2) ⇒sin 2α=(1/2) ⇒2α=30° ⇒α=15°

$${r}={radius} \\ $$$${AD}={DC} \\ $$$$\Rightarrow{CF}=\frac{{OB}}{\mathrm{2}}=\frac{{r}}{\mathrm{2}} \\ $$$${CD}={CE}×\mathrm{cos}\:\alpha=\mathrm{2}{r}\:\mathrm{cos}\:\alpha \\ $$$${CF}={CD}×\mathrm{sin}\:\alpha=\mathrm{2}{r}\:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\alpha={r}\:\mathrm{sin}\:\mathrm{2}\alpha \\ $$$$\Rightarrow{r}\:\mathrm{sin}\:\mathrm{2}\alpha=\frac{{r}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{2}\alpha=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{2}\alpha=\mathrm{30}°\:\Rightarrow\alpha=\mathrm{15}° \\ $$

Commented by M±th+et£s last updated on 09/Apr/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com