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Question-15175

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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=13,AB=14,AC=15 DJ,is the perpendicular bisector of AC. DI⊥BC. ........................ Radius of inescribed circle in triangle DI^Δ J=?
$${in}\:{triangle}\:{ABC}: \\ $$$${BC}=\mathrm{13},{AB}=\mathrm{14},{AC}=\mathrm{15} \\ $$$${DJ},{is}\:{the}\:{perpendicular}\:{bisector}\:{of}\:{AC}. \\ $$$${DI}\bot{BC}. \\ $$$$…………………… \\ $$$${Radius}\:{of}\:{inescribed}\:{circle}\:{in}\:{triangle} \\ $$$${D}\overset{\Delta} {{I}J}=? \\ $$
Answered by mrW1 last updated on 08/Jun/17
cos A=((14^2 +15^2 −13^2 )/(2×14×15))=(3/5) sin A=(√(1−((3/5))^2 ))=(4/5) cos C=((13^2 +15^2 −14^2 )/(2×13×15))=((33)/(65)) sin C=(√(1−(((33)/(65)))^2 ))=((56)/(65)) DJ=((15)/2)×tan A=((15)/2)×(4/3)=10 DI=((15)/2)×sin C=((15)/2)×((56)/(65))=6.46 JI=(√(DJ^2 +DI^2 −2×DJ×DI×cos ∠JDI)) ∠JDI=∠C ⇒JI=(√(10^2 +6.46^2 −2×10×6.46×((33)/(65))))=8.73 s=(10+6.46+8.73)/2=12.59 r_i =(√(((12.59−10)(12.59−6.46)(12.59−8.73))/(12.59)))=2.21
$$\mathrm{cos}\:\mathrm{A}=\frac{\mathrm{14}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} −\mathrm{13}^{\mathrm{2}} }{\mathrm{2}×\mathrm{14}×\mathrm{15}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\mathrm{sin}\:\mathrm{A}=\sqrt{\mathrm{1}−\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} }=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{cos}\:\mathrm{C}=\frac{\mathrm{13}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} −\mathrm{14}^{\mathrm{2}} }{\mathrm{2}×\mathrm{13}×\mathrm{15}}=\frac{\mathrm{33}}{\mathrm{65}} \\ $$$$\mathrm{sin}\:\mathrm{C}=\sqrt{\mathrm{1}−\left(\frac{\mathrm{33}}{\mathrm{65}}\right)^{\mathrm{2}} }=\frac{\mathrm{56}}{\mathrm{65}} \\ $$$$\mathrm{DJ}=\frac{\mathrm{15}}{\mathrm{2}}×\mathrm{tan}\:\mathrm{A}=\frac{\mathrm{15}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{3}}=\mathrm{10} \\ $$$$\mathrm{DI}=\frac{\mathrm{15}}{\mathrm{2}}×\mathrm{sin}\:\mathrm{C}=\frac{\mathrm{15}}{\mathrm{2}}×\frac{\mathrm{56}}{\mathrm{65}}=\mathrm{6}.\mathrm{46} \\ $$$$\mathrm{JI}=\sqrt{\mathrm{DJ}^{\mathrm{2}} +\mathrm{DI}^{\mathrm{2}} −\mathrm{2}×\mathrm{DJ}×\mathrm{DI}×\mathrm{cos}\:\angle\mathrm{JDI}} \\ $$$$\angle\mathrm{JDI}=\angle\mathrm{C} \\ $$$$\Rightarrow\mathrm{JI}=\sqrt{\mathrm{10}^{\mathrm{2}} +\mathrm{6}.\mathrm{46}^{\mathrm{2}} −\mathrm{2}×\mathrm{10}×\mathrm{6}.\mathrm{46}×\frac{\mathrm{33}}{\mathrm{65}}}=\mathrm{8}.\mathrm{73} \\ $$$$\mathrm{s}=\left(\mathrm{10}+\mathrm{6}.\mathrm{46}+\mathrm{8}.\mathrm{73}\right)/\mathrm{2}=\mathrm{12}.\mathrm{59} \\ $$$$\mathrm{r}_{\mathrm{i}} =\sqrt{\frac{\left(\mathrm{12}.\mathrm{59}−\mathrm{10}\right)\left(\mathrm{12}.\mathrm{59}−\mathrm{6}.\mathrm{46}\right)\left(\mathrm{12}.\mathrm{59}−\mathrm{8}.\mathrm{73}\right)}{\mathrm{12}.\mathrm{59}}}=\mathrm{2}.\mathrm{21} \\ $$
Commented by mrW1 last updated on 08/Jun/17
area of a triangle A with sides a,b,c is A=(√(s(a−a)(s−b)(s−c))) A=(1/2)ar_i +(1/2)br_i +(1/2)cr_i =(1/2)(a+b+c)r_i =sr_i ⇒r_i =(√(((s−a)(s−b)(s−c))/s))
$$\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{A}\:\mathrm{with}\:\mathrm{sides}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{is} \\ $$$$\mathrm{A}=\sqrt{\mathrm{s}\left(\mathrm{a}−\mathrm{a}\right)\left(\mathrm{s}−\mathrm{b}\right)\left(\mathrm{s}−\mathrm{c}\right)} \\ $$$$\mathrm{A}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ar}_{\mathrm{i}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{br}_{\mathrm{i}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cr}_{\mathrm{i}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\mathrm{r}_{\mathrm{i}} =\mathrm{sr}_{\mathrm{i}} \\ $$$$\Rightarrow\mathrm{r}_{\mathrm{i}} =\sqrt{\frac{\left(\mathrm{s}−\mathrm{a}\right)\left(\mathrm{s}−\mathrm{b}\right)\left(\mathrm{s}−\mathrm{c}\right)}{\mathrm{s}}} \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 08/Jun/17
thank you dear mrW1. it is perfect.
$${thank}\:{you}\:{dear}\:{mrW}\mathrm{1}.\:{it}\:{is}\:{perfect}. \\ $$$$ \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 09/Jun/17
∡CDI=90−∡C,∡IDJ=∡C. cos(90−C)=((DI)/(b/2))⇒DI=(b/2).sinC tgA=((DJ)/(b/2))⇒DJ=(b/2).tgA h_I =DI.sinC=(b/2).sinC.sinC=(b/2)sin^2 C h_J =DJ.sinC=(b/2).tgA.sinC IJ^2 =DI^2 +DJ^2 −2DI.DJ.cosC= =(b^2 /4)sin^2 C+(b^2 /4)tg^2 A−2(b^2 /4)tgA.sinCcosC ⇒IJ=(b/2)(√(tg^2 A+sin^2 C−tgA.sin2C)) IJ.h_D =DI.h_I ⇒h_D =(((b^2 /4)sin^2 C.tgA)/((b/2)(√(tg^2 A+sin^2 C−tgA.sin2C)))) h_D =(b/2).((sin^2 C.tgA)/( (√(tg^2 A+sin^2 C−tgAsin2C)))) (1/r)=(1/h_I )+(1/h_J )+(1/h_D )= =(2/(bsin^2 C))+(2/(btgA.sinC))+((2(√(tg^2 A+sin^2 C−tgAsin2C)))/(btgA.sin^2 C))= =(2/(bsin^2 CtgA))(tgA+sinC+(√(tg^2 A+sin^2 C−tgAsinC))) ⇒r=((bsin^2 C.tgA)/(2(tgA+sinC+(√(tg^2 A+sin^2 C−tgAsin2C)))))
$$\measuredangle{CDI}=\mathrm{90}−\measuredangle{C},\measuredangle{IDJ}=\measuredangle{C}. \\ $$$${cos}\left(\mathrm{90}−{C}\right)=\frac{{DI}}{{b}/\mathrm{2}}\Rightarrow{DI}=\frac{{b}}{\mathrm{2}}.{sinC} \\ $$$${tgA}=\frac{{DJ}}{{b}/\mathrm{2}}\Rightarrow{DJ}=\frac{{b}}{\mathrm{2}}.{tgA} \\ $$$${h}_{{I}} ={DI}.{sinC}=\frac{{b}}{\mathrm{2}}.{sinC}.{sinC}=\frac{{b}}{\mathrm{2}}{sin}^{\mathrm{2}} {C} \\ $$$${h}_{{J}} ={DJ}.{sinC}=\frac{{b}}{\mathrm{2}}.{tgA}.{sinC} \\ $$$${IJ}^{\mathrm{2}} ={DI}^{\mathrm{2}} +{DJ}^{\mathrm{2}} −\mathrm{2}{DI}.{DJ}.{cosC}= \\ $$$$=\frac{{b}^{\mathrm{2}} }{\mathrm{4}}{sin}^{\mathrm{2}} {C}+\frac{{b}^{\mathrm{2}} }{\mathrm{4}}{tg}^{\mathrm{2}} {A}−\mathrm{2}\frac{{b}^{\mathrm{2}} }{\mathrm{4}}{tgA}.{sinCcosC} \\ $$$$\Rightarrow{IJ}=\frac{{b}}{\mathrm{2}}\sqrt{{tg}^{\mathrm{2}} {A}+{sin}^{\mathrm{2}} {C}−{tgA}.{sin}\mathrm{2}{C}} \\ $$$${IJ}.{h}_{{D}} ={DI}.{h}_{{I}} \Rightarrow{h}_{{D}} =\frac{\frac{{b}^{\mathrm{2}} }{\mathrm{4}}{sin}^{\mathrm{2}} {C}.{tgA}}{\frac{{b}}{\mathrm{2}}\sqrt{{tg}^{\mathrm{2}} {A}+{sin}^{\mathrm{2}} {C}−{tgA}.{sin}\mathrm{2}{C}}} \\ $$$${h}_{{D}} =\frac{{b}}{\mathrm{2}}.\frac{{sin}^{\mathrm{2}} {C}.{tgA}}{\:\sqrt{{tg}^{\mathrm{2}} {A}+{sin}^{\mathrm{2}} {C}−{tgAsin}\mathrm{2}{C}}} \\ $$$$\frac{\mathrm{1}}{{r}}=\frac{\mathrm{1}}{{h}_{{I}} }+\frac{\mathrm{1}}{{h}_{{J}} }+\frac{\mathrm{1}}{{h}_{{D}} }= \\ $$$$=\frac{\mathrm{2}}{{bsin}^{\mathrm{2}} {C}}+\frac{\mathrm{2}}{{btgA}.{sinC}}+\frac{\mathrm{2}\sqrt{{tg}^{\mathrm{2}} {A}+{sin}^{\mathrm{2}} {C}−{tgAsin}\mathrm{2}{C}}}{{btgA}.{sin}^{\mathrm{2}} {C}}= \\ $$$$=\frac{\mathrm{2}}{{bsin}^{\mathrm{2}} {CtgA}}\left({tgA}+{sinC}+\sqrt{{tg}^{\mathrm{2}} {A}+{sin}^{\mathrm{2}} {C}−{tgAsinC}}\right) \\ $$$$\Rightarrow{r}=\frac{{bsin}^{\mathrm{2}} {C}.{tgA}}{\mathrm{2}\left({tgA}+{sinC}+\sqrt{{tg}^{\mathrm{2}} {A}+{sin}^{\mathrm{2}} {C}−{tgAsin}\mathrm{2}{C}}\right)} \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 09/Jun/17
tgA=(4/3),sinC=((56)/(65)),cosC=((33)/(65)),b=15 bsin^2 C.tgA=15×((56^2 )/(65^2 ))×(4/3)=14.84 tg^2 A+sin^2 C−tgAsin2C= =((16)/9)+((56^2 )/(65^2 ))−2×(4/3)×((33×56)/(65^2 ))=1.35 ⇒r=((14.84)/(2((4/3)+((56)/(65))+(√(1.35)))))=((14.84)/(6.72))=2.21 .■
$${tgA}=\frac{\mathrm{4}}{\mathrm{3}},{sinC}=\frac{\mathrm{56}}{\mathrm{65}},{cosC}=\frac{\mathrm{33}}{\mathrm{65}},{b}=\mathrm{15} \\ $$$${bsin}^{\mathrm{2}} {C}.{tgA}=\mathrm{15}×\frac{\mathrm{56}^{\mathrm{2}} }{\mathrm{65}^{\mathrm{2}} }×\frac{\mathrm{4}}{\mathrm{3}}=\mathrm{14}.\mathrm{84} \\ $$$${tg}^{\mathrm{2}} {A}+{sin}^{\mathrm{2}} {C}−{tgAsin}\mathrm{2}{C}= \\ $$$$=\frac{\mathrm{16}}{\mathrm{9}}+\frac{\mathrm{56}^{\mathrm{2}} }{\mathrm{65}^{\mathrm{2}} }−\mathrm{2}×\frac{\mathrm{4}}{\mathrm{3}}×\frac{\mathrm{33}×\mathrm{56}}{\mathrm{65}^{\mathrm{2}} }=\mathrm{1}.\mathrm{35} \\ $$$$\Rightarrow{r}=\frac{\mathrm{14}.\mathrm{84}}{\mathrm{2}\left(\frac{\mathrm{4}}{\mathrm{3}}+\frac{\mathrm{56}}{\mathrm{65}}+\sqrt{\mathrm{1}.\mathrm{35}}\right)}=\frac{\mathrm{14}.\mathrm{84}}{\mathrm{6}.\mathrm{72}}=\mathrm{2}.\mathrm{21}\:.\blacksquare \\ $$
Commented by mrW1 last updated on 09/Jun/17
GOO...D! an other way.
$$\mathbb{GOO}…\mathbb{D}!\:\mathrm{an}\:\mathrm{other}\:\mathrm{way}. \\ $$

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