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Question Number 15212 by ajfour last updated on 08/Jun/17

Commented by ajfour last updated on 08/Jun/17

$$SolutiontoQ.15175$$

Answered by ajfour last updated on 08/Jun/17

$$CD=\frac{b}{2},AD=\frac{b}{2}$$$$\mathrm{\angle }HDI+\mathrm{\angle }CDI=\frac{\pi }{2}$$$$\mathrm{\angle }HDI+\frac{\pi }{2}-\mathrm{\angle }C=\frac{\pi }{2}\Rightarrow \mathrm{\angle }HDI=\mathrm{\angle }C$$$$\mathrm{\angle }HDL=\frac{C}{2}$$$$HD=\frac{b}{2}\sin C\cos C$$$$HI=\frac{b}{2}\sin {}^{2}C$$$$JD=\frac{b}{2}\tan A,JH=JD-HD$$$$JH=\frac{b}{2}(\tan A-\sin C\cos C)$$$$\mathrm{\angle }IJH=\mathit{\alpha },\tan \alpha =\frac{HI}{JH}$$$$\tan \alpha =\frac{\sin {}^{2}C}{\tan A-\sin C\cos C}....\left(i\right)$$$$\mathrm{\angle }KJL=\frac{\alpha }{2}$$$$JK+KD=JD$$$$\frac{\mathit{r}}{\tan \left(\mathit{\alpha }/2\right)}+\frac{\mathit{r}}{\tan \left(\mathit{C}/2\right)}=\frac{\mathit{b}}{2}\tan \mathit{A}$$$$so,$$$$\mathit{r}=\frac{\mathit{b}}{2}\tan \mathit{A}\left[\frac{\tan \left(\mathit{\alpha }/2\right)\tan \left(\mathit{C}/2\right)}{\tan \left(\mathit{\alpha }/2\right)+\tan \left(\mathit{C}/2\right)}\right]$$$$............................$$$$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{225+196-169}{2\times 15\times 14}$$$$\cos A=\frac{3}{5}\Rightarrow \tan A=\frac{4}{3}$$$$\cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{169+225-196}{2\times 13\times 15}$$$$=\frac{33}{65}.$$$$\Rightarrow \sin C=\frac{56}{65}and\tan C=\frac{56}{33}$$$$\tan \alpha =\frac{\sin {}^{2}C}{\tan A-\sin C\cos C}$$$$=\frac{{\left(56/65\right)}^2}{\left(4/3\right)-\left(56/65\right)\left(33/65\right)}$$$$\tan \alpha =0.8285\Rightarrow \alpha =0.692rad$$$$C={\tan }^{-1}\left(56/33\right)=1.0383$$$$\mathit{r}=\frac{\left(\mathit{b}/2\right)\tan \mathit{A}}{[\frac{1}{\tan \left(\alpha /2\right)}+\frac{1}{\tan (C/2}]}$$$$=\frac{\left(\frac{15}{2}\times \frac{4}{3}\right)}{[\frac{1}{\tan (0.346)}+\frac{1}{\tan (0.519)}]}$$$$=\frac{10}{(2.774+1.75)}=2.21units\bullet$$

Commented by mrW1 last updated on 08/Jun/17

$$\mathrm{Tip}$$$$\mathrm{when}\mathrm{we}\mathrm{know}\mathrm{the}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{triangle},$$$$\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{its}\mathrm{inscribed}\mathrm{circle}{\mathrm{r}}_{\mathrm{i}}$$$$\mathrm{can}\mathrm{also}\mathrm{be}\mathrm{calculated}\mathrm{like}\mathrm{this}:$$$$\mathrm{Area}\mathrm{of}\mathrm{triangle}\mathrm{A}=\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}$$$$\mathrm{but}$$$$\mathrm{A}=\frac{1}{2}{\mathrm{ar}}_{\mathrm{i}}+\frac{1}{2}{\mathrm{br}}_{\mathrm{i}}+\frac{1}{2}{\mathrm{cr}}_{\mathrm{i}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}\times {\mathrm{r}}_{\mathrm{i}}={\mathrm{sr}}_{\mathrm{i}}$$$$\Rightarrow {\mathrm{r}}_{\mathrm{i}}=\frac{\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}}{\mathrm{s}}=\sqrt{\frac{(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}{\mathrm{s}}}$$

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

$$hellomyfriend:mrAjfour.$$$$yoursolutionissmartandsimple.$$$$iloveitsomuch.thanksalot.$$

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

$$yessir.wehave:S=p.r.$$$$also:\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}=\frac{1}{r}$$

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