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Question Number 31383 by rahul 19 last updated on 07/Mar/18

Commented by rahul 19 last updated on 07/Mar/18

$$butthisans.iamgettingbyassuming$$$$isoscelestriangle.forgeneraltriangle$$$$howcando?$$

Commented by rahul 19 last updated on 07/Mar/18

$$ans.is(0,{2\tan }^{-1}\frac{5}{4}).$$

Commented by MJS last updated on 07/Mar/18

$$\mathrm{so}\mathrm{one}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{is}$$$$5\mathrm{units}\mathrm{long},{\mathrm{let}}^{\prime }\mathrm{s}\mathrm{call}\mathrm{it}c$$$$c=5$$$$A\mathrm{rea}=\frac{c\times h_c}{2}$$$$5=\frac{5h_c}{2}$$$$h_c=2$$$$\mathrm{so}\mathrm{the}\mathrm{point}C\mathrm{can}\mathrm{be}\mathrm{placed}\mathrm{on}$$$$\mathrm{the}2\mathrm{straight}\mathrm{lines}{g}_1,g_2\mathrm{parallel}$$$$\mathrm{to}\mathrm{our}\mathrm{given}\mathrm{line}c\mathrm{and}\mathrm{with}$$$$\mid g_{1}c\mid =\mid g_{2}c\mid =h_c=2$$

Commented by rahul 19 last updated on 08/Mar/18

$$sohowdoesthisprovesthattriangle$$$$isisosceles?howuget(0,{2\tan }^{-1}\frac{5}{4})??$$

Commented by MJS last updated on 08/Mar/18

$$\mathrm{my}\mathrm{answer}\mathrm{proves}\mathrm{nothing},\mathrm{it}$$$$\mathrm{shows}\mathrm{where}C\mathrm{is}\mathrm{possible}\mathrm{to}\mathrm{get}$$$$\mathrm{an}\mathrm{Area}\mathrm{of}5$$

Commented by rahul 19 last updated on 09/Mar/18

$$ok,thanks!$$

Answered by Tinkutara last updated on 08/Mar/18

Commented by Tinkutara last updated on 08/Mar/18

But many things are yet to think upon: (1) C must be point of intersection of perpendicular bisector of AB and the other parallel line for maximum angle ACB. I can't prove this but by drawing several points on that parallel line I see it should be such located… (2) The other line 4x+3y=23 is not considered… (3) The other root 4/5 is not considered…

Commented by rahul 19 last updated on 08/Mar/18

$$hmmm,anywaysthanks!$$

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