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Question Number 82729 by TawaTawa last updated on 23/Feb/20

Commented by Kunal12588 last updated on 23/Feb/20

$$Inthe△ABC,letGbethecentroid,andletI$$$$bethecenteroftheinscribedcircle.Let\alpha$$$$and\beta betheanglesattheverticesAandB$$$$respectively.SupposethatthesegmentIG\parallel AB$$$$andthat\beta =2{tan}^{-1}\left(\frac{1}{3}\right).Find\alpha .$$

Commented by TawaTawa last updated on 23/Feb/20

$$\mathrm{In}\mathrm{the}\mathrm{triangle}\mathrm{\Delta }\mathrm{ABC},\mathrm{let}\mathrm{G}\mathrm{be}\mathrm{the}\mathrm{centroid},\mathrm{and}\mathrm{let}\mathrm{I}$$$$\mathrm{be}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{the}\mathrm{inscribed}\mathrm{circle}.\mathrm{Let}\alpha \mathrm{and}\beta \mathrm{be}\mathrm{the}$$$$\mathrm{angles}\mathrm{at}\mathrm{the}\mathrm{vertices}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{respectively}.\mathrm{Suppose}\mathrm{that}$$$$\mathrm{the}\mathrm{segment}\mathrm{IG}\mathrm{is}\mathrm{parallel}\mathrm{to}\mathrm{AB}\mathrm{and}\mathrm{that}\beta =2{\tan }^{-1}\left(\frac{1}{3}\right).$$$$\mathrm{Find}\alpha .$$

Answered by mr W last updated on 24/Feb/20

Commented by mr W last updated on 24/Feb/20

$$\beta =2{\tan }^{-1}\frac{1}{3}$$$$\tan \frac{\beta }{2}=\frac{1}{3}$$$$\tan \beta =\frac{2\times \frac{1}{3}}{1-{\left(\frac{1}{3}\right)}^2}=\frac{3}{4}$$$$letradiusofincircle=r=1$$$$BF=\frac{IF}{\tan \frac{\beta }{2}}=\frac{r}{\frac{1}{3}}=3r=3$$$$KF=\frac{IF}{\tan \beta }=\frac{r}{\frac{3}{4}}=\frac{4r}{3}=\frac{4}{3}$$$$BK=BF-KF=3-\frac{4}{3}=\frac{5}{3}$$$$\frac{CK}{BK}=\frac{CG}{EG}=\frac{2}{1}$$$$\Rightarrow CK=2BK=\frac{10}{3}$$$$BC=BK+CK=\frac{5}{3}+\frac{10}{3}=5$$$$FC=BC-BF=5-3=2$$$$\tan \frac{\gamma }{2}=\frac{IF}{FC}=\frac{1}{2}$$$$\tan \gamma =\frac{2\times \frac{1}{2}}{1-{\left(\frac{1}{2}\right)}^2}=\frac{4}{3}=\frac{1}{\tan \beta }$$$$\tan \beta \times \tan \gamma =1$$$$\Rightarrow \beta +\gamma =90°$$$$\alpha =180°-\beta -\gamma$$$$\Rightarrow \alpha =90°$$

Commented by TawaTawa last updated on 24/Feb/20

$$\mathrm{Wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{really}\mathrm{appreciate}\mathrm{your}\mathrm{time}.$$

Commented by TawaTawa last updated on 24/Feb/20

$$\mathrm{Sir}\mathrm{from}\mathrm{the}\mathrm{diagram}.\mathrm{How}\mathrm{is}\mathrm{IK}=\mathrm{BE}$$

Commented by mr W last updated on 24/Feb/20

$$whydoyouthinkIK=BE?$$

Commented by mr W last updated on 24/Feb/20

$$youcancalculateIK=\sqrt{1^2+{\left(\frac{4}{3}\right)}^2}=\frac{5}{3}$$$$andBE=\frac{AB}{2}=\frac{4}{2}=2.thereforeIK\ne BE.$$

Commented by TawaTawa1 last updated on 24/Feb/20

$$\mathrm{sorry}\mathrm{sir},\mathrm{i}\mathrm{mean}\mathrm{parallel}\mathrm{to}\mathrm{BE}.$$

Commented by mr W last updated on 24/Feb/20

$$Itisgiven:IGisparalleltoAB!$$

Commented by TawaTawa1 last updated on 24/Feb/20

$$\mathrm{Oh},\mathrm{thank}\mathrm{you}\mathrm{sir},\mathrm{i}\mathrm{appreciate}.$$

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