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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

$$intriangleABC:$$$$BC=13,AB=14,AC=15$$$$DJ,istheperpendicularbisectorofAC.$$$$DI\mathrm{\perp }BC.$$$$........................$$$$Radiusofinescribedcircleintriangle$$$$D\stackrel{\mathrm{\Delta }}{IJ}=?$$

Answered by mrW1 last updated on 08/Jun/17

$$\cos \mathrm{A}=\frac{{14}^2+{15}^2-{13}^2}{2\times 14\times 15}=\frac{3}{5}$$$$\sin \mathrm{A}=\sqrt{1-{\left(\frac{3}{5}\right)}^2}=\frac{4}{5}$$$$\cos \mathrm{C}=\frac{{13}^2+{15}^2-{14}^2}{2\times 13\times 15}=\frac{33}{65}$$$$\sin \mathrm{C}=\sqrt{1-{\left(\frac{33}{65}\right)}^2}=\frac{56}{65}$$$$\mathrm{DJ}=\frac{15}{2}\times \tan \mathrm{A}=\frac{15}{2}\times \frac{4}{3}=10$$$$\mathrm{DI}=\frac{15}{2}\times \sin \mathrm{C}=\frac{15}{2}\times \frac{56}{65}=6.46$$$$\mathrm{JI}=\sqrt{{\mathrm{DJ}}^2+{\mathrm{DI}}^2-2\times \mathrm{DJ}\times \mathrm{DI}\times \cos \mathrm{\angle }\mathrm{JDI}}$$$$\mathrm{\angle }\mathrm{JDI}=\mathrm{\angle }\mathrm{C}$$$$\Rightarrow \mathrm{JI}=\sqrt{{10}^2+6.{46}^2-2\times 10\times 6.46\times \frac{33}{65}}=8.73$$$$\mathrm{s}=(10+6.46+8.73)/2=12.59$$$${\mathrm{r}}_{\mathrm{i}}=\sqrt{\frac{(12.59-10)(12.59-6.46)(12.59-8.73)}{12.59}}=2.21$$

Commented by mrW1 last updated on 08/Jun/17

$$\mathrm{area}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{A}\mathrm{with}\mathrm{sides}\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{is}$$$$\mathrm{A}=\sqrt{\mathrm{s}(\mathrm{a}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}$$$$\mathrm{A}=\frac{1}{2}{\mathrm{ar}}_{\mathrm{i}}+\frac{1}{2}{\mathrm{br}}_{\mathrm{i}}+\frac{1}{2}{\mathrm{cr}}_{\mathrm{i}}=\frac{1}{2}(\mathrm{a}+\mathrm{b}+\mathrm{c}){\mathrm{r}}_{\mathrm{i}}={\mathrm{sr}}_{\mathrm{i}}$$$$\Rightarrow {\mathrm{r}}_{\mathrm{i}}=\sqrt{\frac{(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}{\mathrm{s}}}$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 08/Jun/17

$$thankyoudearmrW1.itisperfect.$$$$$$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 09/Jun/17

$$\measuredangle CDI=90-\measuredangle C,\measuredangle IDJ=\measuredangle C.$$$$cos(90-C)=\frac{DI}{b/2}\Rightarrow DI=\frac{b}{2}.sinC$$$$tgA=\frac{DJ}{b/2}\Rightarrow DJ=\frac{b}{2}.tgA$$$$h_I=DI.sinC=\frac{b}{2}.sinC.sinC=\frac{b}{2}{sin}^{2}C$$$$h_J=DJ.sinC=\frac{b}{2}.tgA.sinC$$$${IJ}^2={DI}^2+{DJ}^2-2DI.DJ.cosC=$$$$=\frac{b^2}{4}{sin}^{2}C+\frac{b^2}{4}{tg}^{2}A-2\frac{b^2}{4}tgA.sinCcosC$$$$\Rightarrow IJ=\frac{b}{2}\sqrt{{tg}^{2}A+{sin}^{2}C-tgA.sin2C}$$$$IJ.h_D=DI.h_I\Rightarrow h_D=\frac{\frac{b^2}{4}{sin}^{2}C.tgA}{\frac{b}{2}\sqrt{{tg}^{2}A+{sin}^{2}C-tgA.sin2C}}$$$$h_D=\frac{b}{2}.\frac{{sin}^{2}C.tgA}{\sqrt{{tg}^{2}A+{sin}^{2}C-tgAsin2C}}$$$$\frac{1}{r}=\frac{1}{h_I}+\frac{1}{h_J}+\frac{1}{h_D}=$$$$=\frac{2}{{bsin}^{2}C}+\frac{2}{btgA.sinC}+\frac{2\sqrt{{tg}^{2}A+{sin}^{2}C-tgAsin2C}}{btgA.{sin}^{2}C}=$$$$=\frac{2}{{bsin}^{2}CtgA}(tgA+sinC+\sqrt{{tg}^{2}A+{sin}^{2}C-tgAsinC})$$$$\Rightarrow r=\frac{{bsin}^{2}C.tgA}{2(tgA+sinC+\sqrt{{tg}^{2}A+{sin}^{2}C-tgAsin2C})}$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 09/Jun/17

$$tgA=\frac{4}{3},sinC=\frac{56}{65},cosC=\frac{33}{65},b=15$$$${bsin}^{2}C.tgA=15\times \frac{{56}^2}{{65}^2}\times \frac{4}{3}=14.84$$$${tg}^{2}A+{sin}^{2}C-tgAsin2C=$$$$=\frac{16}{9}+\frac{{56}^2}{{65}^2}-2\times \frac{4}{3}\times \frac{33\times 56}{{65}^2}=1.35$$$$\Rightarrow r=\frac{14.84}{2(\frac{4}{3}+\frac{56}{65}+\sqrt{1.35})}=\frac{14.84}{6.72}=2.21.◼$$

Commented by mrW1 last updated on 09/Jun/17

$$\mathbb{GOO}...\mathbb{D}!\mathrm{an}\mathrm{other}\mathrm{way}.$$

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