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Question Number 30862 by ajfour last updated on 27/Feb/18

Answered by mrW2 last updated on 27/Feb/18

$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=\frac{1}{\lambda }=2R=\frac{abc}{2\sqrt{s(s-a)(s-b)(s-c)}}$$$$\frac{BQ}{\sin (C-\theta )}=\frac{CQ}{\sin \theta }=\frac{BC}{\sin (\pi -\theta -C+\theta )}$$$$\frac{BQ}{\sin (C-\theta )}=\frac{CQ}{\sin \theta }=\frac{a}{\sin C}$$$$\Rightarrow BQ=\frac{\sin (C-\theta )}{\sin C}\times a$$$$\Rightarrow CQ=\frac{\sin \theta }{\sin C}\times a$$$$similarly$$$$\Rightarrow BP=\frac{\sin \theta }{\sin B}\times c$$$$\Rightarrow a^{\prime }=PQ=BQ-BP=\frac{\sin (C-\theta )}{\sin C}\times a-\frac{\sin \theta }{\sin B}\times c$$$$\Rightarrow a^{\prime }=\frac{c\lambda \cos \theta -\cos C\sin \theta }{c\lambda }\times a-\frac{\sin \theta }{b\lambda }\times c$$$$\Rightarrow a^{\prime }=a\cos \theta -\frac{(ab\cos C+c^2)\sin \theta }{bc\lambda }$$$$\Rightarrow a^{\prime }=a\cos \theta -\frac{(\frac{a^2+b^2-c^2}{2}+c^2)\sin \theta }{bc\lambda }$$$$\Rightarrow a^{\prime }=a\cos \theta -\frac{(a^2+b^2+c^2)\sin \theta }{2bc\lambda }$$$$\Rightarrow a^{\prime }=[\cos \theta -\frac{(a^2+b^2+c^2)\sin \theta }{4\sqrt{s(s-a)(s-b)(s-c)}}]a=ka$$$$similarly$$$$\Rightarrow b^{\prime }=kb$$$$\Rightarrow c^{\prime }=kc$$$$\Rightarrow r=k{r}_{\mathrm{\Delta }ABC}=k\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s}$$$$\Rightarrow r=[\cos \theta -\frac{(a^2+b^2+c^2)\sin \theta }{4\sqrt{s(s-a)(s-b)(s-c)}}]\times \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$$

Commented by ajfour last updated on 27/Feb/18

$$ExcellentSir,ihavejustchecked$$$$itfor\theta =30°witha=b=c.$$$$r=0asitshouldbe.$$$$Thankyousir.\lambda isquitehelpful.$$

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