Question Number 15175 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 08/Jun/17 | ||
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 08/Jun/17 | ||
$${in}\:{triangle}\:{ABC}: \\ $$$${BC}=\mathrm{13},{AB}=\mathrm{14},{AC}=\mathrm{15} \\ $$$${DJ},{is}\:{the}\:{perpendicular}\:{bisector}\:{of}\:{AC}. \\ $$$${DI}\bot{BC}. \\ $$$$........................ \\ $$$${Radius}\:{of}\:{inescribed}\:{circle}\:{in}\:{triangle} \\ $$$${D}\overset{\Delta} {{I}J}=? \\ $$ | ||
Answered by mrW1 last updated on 08/Jun/17 | ||
$$\mathrm{cos}\:\mathrm{A}=\frac{\mathrm{14}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} −\mathrm{13}^{\mathrm{2}} }{\mathrm{2}×\mathrm{14}×\mathrm{15}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\mathrm{sin}\:\mathrm{A}=\sqrt{\mathrm{1}−\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} }=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{cos}\:\mathrm{C}=\frac{\mathrm{13}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} −\mathrm{14}^{\mathrm{2}} }{\mathrm{2}×\mathrm{13}×\mathrm{15}}=\frac{\mathrm{33}}{\mathrm{65}} \\ $$$$\mathrm{sin}\:\mathrm{C}=\sqrt{\mathrm{1}−\left(\frac{\mathrm{33}}{\mathrm{65}}\right)^{\mathrm{2}} }=\frac{\mathrm{56}}{\mathrm{65}} \\ $$$$\mathrm{DJ}=\frac{\mathrm{15}}{\mathrm{2}}×\mathrm{tan}\:\mathrm{A}=\frac{\mathrm{15}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{3}}=\mathrm{10} \\ $$$$\mathrm{DI}=\frac{\mathrm{15}}{\mathrm{2}}×\mathrm{sin}\:\mathrm{C}=\frac{\mathrm{15}}{\mathrm{2}}×\frac{\mathrm{56}}{\mathrm{65}}=\mathrm{6}.\mathrm{46} \\ $$$$\mathrm{JI}=\sqrt{\mathrm{DJ}^{\mathrm{2}} +\mathrm{DI}^{\mathrm{2}} −\mathrm{2}×\mathrm{DJ}×\mathrm{DI}×\mathrm{cos}\:\angle\mathrm{JDI}} \\ $$$$\angle\mathrm{JDI}=\angle\mathrm{C} \\ $$$$\Rightarrow\mathrm{JI}=\sqrt{\mathrm{10}^{\mathrm{2}} +\mathrm{6}.\mathrm{46}^{\mathrm{2}} −\mathrm{2}×\mathrm{10}×\mathrm{6}.\mathrm{46}×\frac{\mathrm{33}}{\mathrm{65}}}=\mathrm{8}.\mathrm{73} \\ $$$$\mathrm{s}=\left(\mathrm{10}+\mathrm{6}.\mathrm{46}+\mathrm{8}.\mathrm{73}\right)/\mathrm{2}=\mathrm{12}.\mathrm{59} \\ $$$$\mathrm{r}_{\mathrm{i}} =\sqrt{\frac{\left(\mathrm{12}.\mathrm{59}−\mathrm{10}\right)\left(\mathrm{12}.\mathrm{59}−\mathrm{6}.\mathrm{46}\right)\left(\mathrm{12}.\mathrm{59}−\mathrm{8}.\mathrm{73}\right)}{\mathrm{12}.\mathrm{59}}}=\mathrm{2}.\mathrm{21} \\ $$ | ||
Commented by mrW1 last updated on 08/Jun/17 | ||
$$\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{A}\:\mathrm{with}\:\mathrm{sides}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{is} \\ $$$$\mathrm{A}=\sqrt{\mathrm{s}\left(\mathrm{a}−\mathrm{a}\right)\left(\mathrm{s}−\mathrm{b}\right)\left(\mathrm{s}−\mathrm{c}\right)} \\ $$$$\mathrm{A}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ar}_{\mathrm{i}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{br}_{\mathrm{i}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cr}_{\mathrm{i}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\mathrm{r}_{\mathrm{i}} =\mathrm{sr}_{\mathrm{i}} \\ $$$$\Rightarrow\mathrm{r}_{\mathrm{i}} =\sqrt{\frac{\left(\mathrm{s}−\mathrm{a}\right)\left(\mathrm{s}−\mathrm{b}\right)\left(\mathrm{s}−\mathrm{c}\right)}{\mathrm{s}}} \\ $$ | ||
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 08/Jun/17 | ||
$${thank}\:{you}\:{dear}\:{mrW}\mathrm{1}.\:{it}\:{is}\:{perfect}. \\ $$$$ \\ $$ | ||
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 09/Jun/17 | ||
$$\measuredangle{CDI}=\mathrm{90}−\measuredangle{C},\measuredangle{IDJ}=\measuredangle{C}. \\ $$$${cos}\left(\mathrm{90}−{C}\right)=\frac{{DI}}{{b}/\mathrm{2}}\Rightarrow{DI}=\frac{{b}}{\mathrm{2}}.{sinC} \\ $$$${tgA}=\frac{{DJ}}{{b}/\mathrm{2}}\Rightarrow{DJ}=\frac{{b}}{\mathrm{2}}.{tgA} \\ $$$${h}_{{I}} ={DI}.{sinC}=\frac{{b}}{\mathrm{2}}.{sinC}.{sinC}=\frac{{b}}{\mathrm{2}}{sin}^{\mathrm{2}} {C} \\ $$$${h}_{{J}} ={DJ}.{sinC}=\frac{{b}}{\mathrm{2}}.{tgA}.{sinC} \\ $$$${IJ}^{\mathrm{2}} ={DI}^{\mathrm{2}} +{DJ}^{\mathrm{2}} −\mathrm{2}{DI}.{DJ}.{cosC}= \\ $$$$=\frac{{b}^{\mathrm{2}} }{\mathrm{4}}{sin}^{\mathrm{2}} {C}+\frac{{b}^{\mathrm{2}} }{\mathrm{4}}{tg}^{\mathrm{2}} {A}−\mathrm{2}\frac{{b}^{\mathrm{2}} }{\mathrm{4}}{tgA}.{sinCcosC} \\ $$$$\Rightarrow{IJ}=\frac{{b}}{\mathrm{2}}\sqrt{{tg}^{\mathrm{2}} {A}+{sin}^{\mathrm{2}} {C}−{tgA}.{sin}\mathrm{2}{C}} \\ $$$${IJ}.{h}_{{D}} ={DI}.{h}_{{I}} \Rightarrow{h}_{{D}} =\frac{\frac{{b}^{\mathrm{2}} }{\mathrm{4}}{sin}^{\mathrm{2}} {C}.{tgA}}{\frac{{b}}{\mathrm{2}}\sqrt{{tg}^{\mathrm{2}} {A}+{sin}^{\mathrm{2}} {C}−{tgA}.{sin}\mathrm{2}{C}}} \\ $$$${h}_{{D}} =\frac{{b}}{\mathrm{2}}.\frac{{sin}^{\mathrm{2}} {C}.{tgA}}{\sqrt{{tg}^{\mathrm{2}} {A}+{sin}^{\mathrm{2}} {C}−{tgAsin}\mathrm{2}{C}}} \\ $$$$\frac{\mathrm{1}}{{r}}=\frac{\mathrm{1}}{{h}_{{I}} }+\frac{\mathrm{1}}{{h}_{{J}} }+\frac{\mathrm{1}}{{h}_{{D}} }= \\ $$$$=\frac{\mathrm{2}}{{bsin}^{\mathrm{2}} {C}}+\frac{\mathrm{2}}{{btgA}.{sinC}}+\frac{\mathrm{2}\sqrt{{tg}^{\mathrm{2}} {A}+{sin}^{\mathrm{2}} {C}−{tgAsin}\mathrm{2}{C}}}{{btgA}.{sin}^{\mathrm{2}} {C}}= \\ $$$$=\frac{\mathrm{2}}{{bsin}^{\mathrm{2}} {CtgA}}\left({tgA}+{sinC}+\sqrt{{tg}^{\mathrm{2}} {A}+{sin}^{\mathrm{2}} {C}−{tgAsinC}}\right) \\ $$$$\Rightarrow{r}=\frac{{bsin}^{\mathrm{2}} {C}.{tgA}}{\mathrm{2}\left({tgA}+{sinC}+\sqrt{{tg}^{\mathrm{2}} {A}+{sin}^{\mathrm{2}} {C}−{tgAsin}\mathrm{2}{C}}\right)} \\ $$ | ||
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 09/Jun/17 | ||
$${tgA}=\frac{\mathrm{4}}{\mathrm{3}},{sinC}=\frac{\mathrm{56}}{\mathrm{65}},{cosC}=\frac{\mathrm{33}}{\mathrm{65}},{b}=\mathrm{15} \\ $$$${bsin}^{\mathrm{2}} {C}.{tgA}=\mathrm{15}×\frac{\mathrm{56}^{\mathrm{2}} }{\mathrm{65}^{\mathrm{2}} }×\frac{\mathrm{4}}{\mathrm{3}}=\mathrm{14}.\mathrm{84} \\ $$$${tg}^{\mathrm{2}} {A}+{sin}^{\mathrm{2}} {C}−{tgAsin}\mathrm{2}{C}= \\ $$$$=\frac{\mathrm{16}}{\mathrm{9}}+\frac{\mathrm{56}^{\mathrm{2}} }{\mathrm{65}^{\mathrm{2}} }−\mathrm{2}×\frac{\mathrm{4}}{\mathrm{3}}×\frac{\mathrm{33}×\mathrm{56}}{\mathrm{65}^{\mathrm{2}} }=\mathrm{1}.\mathrm{35} \\ $$$$\Rightarrow{r}=\frac{\mathrm{14}.\mathrm{84}}{\mathrm{2}\left(\frac{\mathrm{4}}{\mathrm{3}}+\frac{\mathrm{56}}{\mathrm{65}}+\sqrt{\mathrm{1}.\mathrm{35}}\right)}=\frac{\mathrm{14}.\mathrm{84}}{\mathrm{6}.\mathrm{72}}=\mathrm{2}.\mathrm{21}\:.\blacksquare \\ $$ | ||
Commented by mrW1 last updated on 09/Jun/17 | ||
$$\mathbb{GOO}...\mathbb{D}!\:\mathrm{an}\:\mathrm{other}\:\mathrm{way}. \\ $$ | ||